While trying to develop the generating function for \begin{align} a_{n} = \begin{cases} 1 & n=1 \\ \sum_{k=0}^{n-1} (-1)^k \, \binom{n+1}{k} \, \frac{(n-1)!}{(n+1) \, (n-k-1)!} \, (n-k+1)^{n-k-1} & n \geq 2 \end{cases} \end{align} some problems arise.
The generating function begins as the following. \begin{align} \sum_{n=1}^{\infty} a_{n} \, \frac{t^{n}}{n!} &= t + \sum_{n=2}^{\infty} \sum_{k=0}^{n-1} (-1)^k \, \binom{n+1}{k} \, \frac{(n-1)!}{(n+1) \, (n-k-1)!} \, (n-k+1)^{n-k-1} \, \frac{t^{n}}{n!} \\ &= t + \sum_{n=2}^{\infty} \sum_{k=0}^{n-1} \frac{(-1)^k \, (n-1)!}{k! \, (n-k+1)! \, (n-k-1)!} \, (n-k+1)^{n-k-1} \, t^{n} \\ &= t + \sum_{n=1}^{\infty} \sum_{k=0}^{n} \frac{(-1)^k \, n!}{k! \, (n-k+2)! \, (n-k)!} \, (n-k+2)^{n-k} \, t^{n+1} \\ &= \frac{t}{2} + \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^k \, n!}{k! \, (n-k+2)! \, (n-k)!} \, (n-k+2)^{n-k} \, t^{n+1} \\ &= \frac{t}{2} + \sum_{n,k=0}^{\infty} \frac{(-1)^k \, (n+k)!}{k! \, (n+2)! \, n!} \, (n+2)^{n} \, t^{n+k+1} \\ &= \frac{t}{2} + \sum_{n=0}^{\infty} \frac{(n+2)^{n-1} \, t^{n+1}}{(n+2)!} \, \sum_{k=0}^{\infty} \binom{n+k}{k} \, (-t)^{k} \\ &= \frac{t}{2} + \sum_{n=0}^{\infty} \frac{(n+2)^{(n+2)-3}}{(n+2)!} \, \left(\frac{t}{1+t}\right)^{n+1} \\ &= \frac{t}{2} - \frac{1+t}{t} \, \sum_{n=2}^{\infty} \frac{(-n)^{n-3}}{n!} \, \left(\frac{-t}{1+t}\right)^{n}. \end{align}
At this point use of $$\frac{1}{n (n-1)} = \int_{0}^{1} u^{n-2} \, (1-u) \, du$$ can be made to yield $$ \sum_{n=1}^{\infty} a_{n} \frac{t^{n}}{n!} = \frac{t}{2} + \frac{1+t}{2 \, t} \, \int_{0}^{1} W^{2}\left(\frac{-t u}{1+t}\right) \, \frac{1 - u}{u^{2}} \, du,$$ where $W(x)$ is the Lambert W-function.
The Question: How can this result be reduced? And if it can be reduced can it match the believed result of $$ \sum_{n=1}^{\infty} a_{n} \frac{t^{n}}{n!} = \frac{1}{2 \, t} - \frac{1 - t}{2} - \frac{1 + t}{2 \, t} \, \left( 1 + W\left(\frac{-t}{1+t}\right) \right)^{2} \quad?$$
I do not know how much this could help you $$\int_0^1 W^2(a u)\frac{(1-u) }{u^2}\,du=2a-2 W(a)-\frac{3 }{2}W^2(a)-\frac{1}{3} W^3(a)$$
So, they do not seem to match.
Using Taylor series around $t=0$, the penultimate expression would give $\frac{t}{2}+\frac{t^3}{4}+O\left(t^4\right)$ while the last would give $t+\frac{t^3}{6}+O\left(t^4\right)$.