Proving $\sum\limits_{k=0}^n \sum\limits_{j=0}^{n-k} \frac{(k-1)^2}{k!} \frac{(-1)^j}{j!} =1$ without character theory

We may start from: $$ \sum_{k\geq 0} \frac{(k-1)^2}{k!}x^k = (1-x+x^2)\,e^{x}\tag{1} $$ $$ \sum_{k\geq 0}\left(\sum_{j=0}^{k}\frac{(-1)^j}{j!}\right) x^k = \frac{e^{-x}}{1-x}\tag{2} $$ then notice that the original sum is just the coefficient of $x^n$ in the product between the RHSs of $(1)$ and $(2)$, i.e.

$$ [x^n]\left(\frac{1-x+x^2}{1-x}\right)=[x^n]\left(-x+\color{red}{\frac{1}{1-x}}\right)\tag{3}$$

Now the claim is pretty trivial: the original sum equals $1$ for every $n\in\mathbb{N}$, except for $n=1$ where it equals zero.