You assume that exponentation of real numbers by real numbers satisfies $a^{p \cdot q}= (a^p)^q$. However, it is not that simple. It is true for any $a \in \mathbb R$ and any $p,q \in \mathbb N$. But what is $a^x$ for a non-integer $x$? For $a > 0$ there are various approaches to define it.

(a) $a^x = e^{x \ln a}$ for all $x \in \mathbb R$.

(b) $a^{r/s} = \sqrt[s]{a^r}$ for all $r/s \in \mathbb Q$ where we understand $s \in \mathbb N$.

The second approach can be used to define $a^x$ as $\lim_{r/s \to x} a^{r/s}$, but this requires some work.

For $a >0$ both approaches yield $a^{x \cdot y}= (a^x)^y$ for $x,y \in \mathbb R$ resp. $x,y\in \mathbb Q$.

For $a < 0$ we get troubles. The first approach fails beacuse $\ln a$ is not defined (as a real number). The second approach has serious problems:

(1) It can only work when $r$ is even or $s$ is odd, otherwise you get something undefined (at least if you want a real value for $a^{r/s}$).

(2) If both $r,s$ are even, then the $s$-th root has two possible values (a positive and a negative). You may think that we should always choose the positive value, but the consequences would be unpleasant as you will in the next point.

(3) We should expect that $a^{r/s} = a^{u/v}$ if $r/s = u/v$. But if both $r, s$ are odd, then $a^{r/s}$ is negative whereas $a^{2r/2s}$ is positive.

Choosing always the negative value for the $s$-th root, $s$ even, produces the same problem (consider $r$ even, $s$ odd). And choosing in an ad-hoc way cannot be a serious approach.

Thus, if $a < 0$, you cannot expect $a^{x \cdot y}= (a^x)^y$ to be true no matter how you define $a^{r/s}$. In your question you gave an example:

$$((-1)^2)^{1/2} = 1^{1/2}= \sqrt{1} = 1 \ne (-1)^{2 \cdot 1/2} = (-1)^1 = -1$$ if we choose the positive root.

The lesson is: Be careful when using $a^{x \cdot y}= (a^x)^y$.

You assume that exponentation of real numbers by real numbers satisfies $a^{p \cdot q}= (a^p)^q$. However, it is not that simple. It is true for any $a \in \mathbb R$ and any $p,q \in \mathbb N$. But what is $a^x$ for a non-integer $x$? For $a > 0$ there are various approaches to define it.

(a) $a^x = e^{x \ln a}$ for all $x \in \mathbb R$.

(b) $a^{r/s} = \sqrt[s]{a^r}$ for all $r/s \in \mathbb Q$ where we understand $s \in \mathbb N$.

The second approach can be used to define $a^x$ as $\lim_{r/s \to x} a^{r/s}$, but this requires some work.

For $a >0$ both approaches yield $a^{x \cdot y}= (a^x)^y$ for $x,y \in \mathbb R$ resp. $x,y\in \mathbb Q$.

For $a < 0$ we get troubles. The first approach fails beacuse $\ln a$ is not defined (as a real number). The second approach has serious problems:

(1) It can only work when $r$ is even or $s$ is odd, otherwise you get something undefined (at least if you want a real value for $a^{r/s}$).

(2) If both $r,s$ are even, then the $s$-th root has two possible values (a positive and a negative). You may think that we should always choose the positive value, but the consequences would be unpleasant as you will in the next point.

(3) We should expect that $a^{r/s} = a^{u/v}$ if $r/s = u/v$. But if both $r, s$ are odd, then $a^{r/s}$ is negative whereas $a^{2r/2s}$ is positive.

Choosing always the negative value for the $s$-th root, $s$ even, produces the same problem (consider $r$ even, $s$ odd). And choosing in an ad-hoc way cannot be a serious approach.

Thus, if $a < 0$, you cannot expect $a^{x \cdot y}= (a^x)^y$ to be true no matter how you define $a^{r/s}$. Here is an example, similar to your question:

$$((-1)^2)^{1/2} = 1^{1/2}= \sqrt{1} = 1 \ne (-1)^{2 \cdot 1/2} = (-1)^1 = -1$$ if we choose the positive root.

The lesson is: Be careful when using $a^{x \cdot y}= (a^x)^y$.

You assume that exponentation of real numbers by real numbers satisfies $a^{p \cdot q}= (a^p)^q$. However, it is not that simple. It is true for any $a \in \mathbb R$ and any $p,q \in \mathbb N$. But what is $a^x$ for a non-integer $x$? For $a > 0$ there are various approaches to define it.

(a) $a^x = e^{x \ln a}$ for all $x \in \mathbb R$.

(b) $a^{r/s} = \sqrt[s]{a^r}$ for all $r/s \in \mathbb Q$ where we understand $s \in \mathbb N$.

The second approach can be used to define $a^x$ as $\lim_{r/s \to x} a^{r/s}$, but this requires some work.

For $a >0$ both approaches yield $a^{x \cdot y}= (a^x)^y$ for $x,y \in \mathbb R$ resp. $x,y\in \mathbb Q$.

For $a < 0$ we get troubles. The first approach fails beacuse $\ln a$ is not defined (as a real number). The second approach has two problems: It can only work when $r$ is even or $s$ is odd, and if both $r,s$ are even, then the $s$-th root has two possible values (a positive and a negative). You may think that we should always choose the positive value, but the consequence is that $a^{x \cdot y}= (a^x)^y$ is in general no longer true even if the three rationals $x,y, x\cdot y$ are as required above. In your question you gave an example:

$$((-1)^2)^{1/2} = 1^{1/2}= \sqrt{1} = 1 \ne (-1)^{2 \cdot 1/2} = (-1)^1 = -1$$ if we choose the positive root.

The lesson is: Be careful when using $a^{x \cdot y}= (a^x)^y$.

You assume that exponentation of real numbers by real numbers satisfies $a^{p \cdot q}= (a^p)^q$. However, it is not that simple. It is true for any $a \in \mathbb R$ and any $p,q \in \mathbb N$. But what is $a^x$ for a non-integer $x$? For $a > 0$ there are various approaches to define it.

(a) $a^x = e^{x \ln a}$ for all $x \in \mathbb R$.

(b) $a^{r/s} = \sqrt[s]{a^r}$ for all $r/s \in \mathbb Q$ where we understand $s \in \mathbb N$.

The second approach can be used to define $a^x$ as $\lim_{r/s \to x} a^{r/s}$, but this requires some work.

For $a >0$ both approaches yield $a^{x \cdot y}= (a^x)^y$ for $x,y \in \mathbb R$ resp. $x,y\in \mathbb Q$.

For $a < 0$ we get troubles. The first approach fails beacuse $\ln a$ is not defined (as a real number). The second approach has serious problems:

(1) It can only work when $r$ is even or $s$ is odd, otherwise you get something undefined (at least if you want a real value for $a^{r/s}$).

(2) If both $r,s$ are even, then the $s$-th root has two possible values (a positive and a negative). You may think that we should always choose the positive value, but the consequences would be unpleasant as you will in the next point.

(3) We should expect that $a^{r/s} = a^{u/v}$ if $r/s = u/v$. But if both $r, s$ are odd, then $a^{r/s}$ is negative whereas $a^{2r/2s}$ is positive.

Choosing always the negative value for the $s$-th root, $s$ even, produces the same problem (consider $r$ even, $s$ odd). And choosing in an ad-hoc way cannot be a serious approach.

Thus, if $a < 0$, you cannot expect $a^{x \cdot y}= (a^x)^y$ to be true no matter how you define $a^{r/s}$. In your question you gave an example:

$$((-1)^2)^{1/2} = 1^{1/2}= \sqrt{1} = 1 \ne (-1)^{2 \cdot 1/2} = (-1)^1 = -1$$ if we choose the positive root.

The lesson is: Be careful when using $a^{x \cdot y}= (a^x)^y$.