Integer coefficients defined recursively with main diagonal equals A195979

Let

• $A_n(x)$ be the family of exponential generating fucntions such that $$ A'_n(x) = nA_n(x) + A_{n-1}(x), \\ A_n(0) = 1, A_0(x) = 1. $$
• $T(n,k)$ be an integer coefficients whose exponential generating function is $A_n(x)$. Obviously we have $$ T(n,k) = nT(n,k-1) + T(n-1,k-1), \\ T(n,0) = 1, T(0,k) = [k=0]. $$ Here square bracket denotes Iverson bracket.
• $a(n)$ be A195979, i.e., an integer sequence whose exponential function satisfies $$ \frac{\exp(1-\exp(W(-x)))}{1+W(-x)}. $$ Here $W(x)$ denotes Lambert W function. Also we have $$ a(n) = 1 + \sum\limits_{k=0}^{n-1} \sum\limits_{m=k}^{n} (m-k) \binom{n+1}{m+1} {m+1 \brace k+1} (n-k)^{n-m-1}. $$

I conjecture that $a(n) = T(n,n)$.

Here is the PARI/GP program to check it numerically:

a(n) = {1 + sum(k=0, n-1, sum(m=k, n, 
(m-k)*binomial(n+1,m+1)*stirling(m+1,k+1,2)*(n-k)^(n-m-1)))}
upto1(n) = {my(x = 'x + 'x*O('x^n)); 
Vec(serlaplace(exp(1-exp(lambertw(-x)))/(1+lambertw(-x))))}
upto2(n) = {my(v1, v2, v3); 
v1 = vector(n+1, i, 0); 
v1[1] = 1; v2 = v1; 
for(i=1, n, v3 = v1; 
for(j=1, i, v1[j+1] = i*v1[j] + v3[j]); 
v2[i+1] = v1[i+1]); v2}
test1(n) = vector(n+1, i, a(i-1)) == upto2(n)
test2(n) = upto1(n) == upto2(n)

Is there a way to prove it? Is there a way to find a solution for $A_n(x)$?