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 
Let $n \geq 2$ be an integer. I would like to prove the following identity in an easy way:
  $$\sum\limits_{k=0}^n \left( \frac{(k-1)^2}{k!} \sum\limits_{j=0}^{n-k} \frac{(-1)^j}{j!} \right)=1$$

You can see on WolframAlpha, for $n=3$, that this holds.

Here is nice proof I've found. However it seems to be awfully sophisticated...
Consider the representation $S_n \to \text{GL}(V)$ where $V=\text{span}_{\Bbb C}(e_1-e_2, \dots, e_1-e_n)$ and $\sigma \cdot (e_1-e_i) := e_{\sigma(1)}- e_{\sigma(i)}\,$.
It is well-known that this is an irrep: we can show that if $0\neq u \in V$, then $\text{span}_{\Bbb C}(\{\sigma \cdot u \mid \sigma \in S_n\}) = V$.
In particular, the scalar product of character is
$\langle \chi_{V}, \chi_V \rangle=1$.
But $\chi_V = \chi_{S_n} - \chi_1$ where $\chi_{S_n} $ is the permutation character of $S_n$ and $\chi_{1}$ the trivial character. 
$\newcommand{\supp}{\text{supp}}$
Therefore, if $\supp(\sigma)$ denotes the support of a permutation $\sigma$, we have:
$$ \begin{align}
\langle \chi_{V}, \chi_V \rangle &=
\frac{1}{n!} \sum\limits_{\sigma \in S_n} |\chi_{S_n}(\sigma)-1|^2\\&=
\frac{1}{n!} \sum\limits_{\sigma \in S_n}
\left|\text{card}(\{1, \dots, n\} \setminus \supp(\sigma)) \;-\; 1\right|^2\\&=
\frac{1}{n!} \sum\limits_{m=0}^n\; \sum\limits_{\substack{\sigma \in S_n\\ |\supp(\sigma)|=m}}
\left|n-m \;-\; 1\right|^2\\&\stackrel{(*)}{=}
\frac{1}{n!} \sum\limits_{m=0}^n (n-m- 1)^2 {n \choose m} m! \sum\limits_{j=0}^m  \frac{(-1)^j}{j!} \\&=
\sum\limits_{m=0}^n (n-m- 1)^2\frac{1}{(n-m)!} \sum\limits_{j=0}^m  \frac{(-1)^j}{j!} \\&=
\sum\limits_{k=0}^n \left( \frac{(k-1)^2}{k!} \sum\limits_{j=0}^{n-k} \frac{(-1)^j}{j!} \right)=1
 \end{align} $$
The equality $(*)$ holds because to choose $\sigma \in S_n$ has a support of cardinality $m$, I choose $m$ numbers out of $n$, and then I have $m!\sum\limits_{j=0}^{m} \frac{(-1)^j}{j!}$ derangements of these $m$ elements.

The following posts are close to my identity, but different: (1), (2). This one only proves $\sum\limits_{k=0}^n \left( \frac{1}{k!} \sum\limits_{j=0}^{n-k} \frac{(-1)^j}{j!} \right)=1$.
In all cases, I would like to see some easy proofs of my identity.
Thank you for your comments!

Another method is induction. 
$f(n):=\sum\limits_{k=0}^n \left( \frac{(k-1)^2}{k!} \sum\limits_{j=0}^{n-k} \frac{(-1)^j}{j!} \right)$  
It's easy to calculate $f(n+1)-f(n)=0$ for $n\ge 2$:
$f(0)=1$, $f(1)=0$, $f(2)=1$
$$f(n+1)-f(n)=
\sum\limits_{k=0}^{n+1}\frac{(k-1)^2(-1)^{n+1-k}}{k!(n+1-k)!}=\\\sum\limits_{k=2}^{n+1}\frac{k(k-1)(-1)^{n+1-k}}{k!(n+1-k)!} - \sum\limits_{k=1}^{n+1}\frac{k(-1)^{n+1-k}}{k!(n+1-k)!}  +\sum\limits_{k=0}^{n+1}\frac{(-1)^{n+1-k}}{k!(n+1-k)!}=\\
\frac{0^{n-1}}{(n-1)!}-\frac{0^n}{n!}+\frac{0^{n+1}}{(n+1)!}=0$$
for $n\ge 2$
