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!
 A: 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.
A: 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 $n\ge 2$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{\sum_{k = 0}^{n}\bracks{{\pars{k - 1}^{2} \over k!}
\sum_{j = 0}^{n - k}{\pars{-1}^{j} \over j!}} = 1:\,?}$

Lets $\ds{a_{k} \equiv {\pars{k - 1}^{2} \over k!}\,,\ b_{j} \equiv {\pars{-1}^{j} \over j!}\ \mbox{such that}\
\sum_{k = 0}^{n}{\pars{k - 1}^{2} \over k!}
\sum_{j = 0}^{n - k}{\pars{-1}^{j} \over j!} =
\sum_{k = 0}^{n}a_{k}\sum_{j = 0}^{n - k}b_{j}}$.

Then, we'll compare the sums for upper limits of $\ds{n}$ and $\ds{n + 1}$:
\begin{align}
\sum_{k = 0}^{n + 1}a_{k}\sum_{j = 0}^{n + 1 - k}b_{j} & =
\sum_{k = 0}^{n}a_{k}\sum_{j = 0}^{n + 1 - k}b_{j} +
a_{n + 1}\sum_{j = 0}^{0}b_{j} =
\sum_{k = 0}^{n}a_{k}\pars{\sum_{j = 0}^{n - k}b_{j} + b_{n + 1 - k}} +
a_{n + 1}
\\[3mm] & =
\sum_{k = 0}^{n}a_{k}\sum_{j = 0}^{n - k}b_{j} +
\sum_{k = 0}^{n}a_{k}b_{n + 1 -k} + a_{n + 1} =
\sum_{k = 0}^{n}a_{k}\sum_{j = 0}^{n - k}b_{j} +
\sum_{k = 0}^{n + 1}a_{k}b_{n + 1 - k}
\end{align}

which leads to
\begin{equation}
\sum_{k = 0}^{n + 1}a_{k}\sum_{j = 0}^{n + 1 - k}b_{j} -
\sum_{k = 0}^{n}a_{k}\sum_{j = 0}^{n - k}b_{j} =
\sum_{k = 0}^{n + 1}a_{k}b_{n + 1 - k}\tag{1}
\end{equation}

In the following step, we'll show that the difference between the sums for the upper limits of $\ds{n + 1}$ and $\ds{n}$ $\pars{~\mbox{the RHS of}\ \pars{1}~}$
vanishes out. Namely,
\begin{align}
\sum_{k = 0}^{n + 1}a_{k}b_{n + 1 -k} & =
\sum_{k = 0}^{n + 1}{\pars{k - 1}^{2} \over k!}\,
{\pars{-1}^{n + 1 - k} \over \pars{n + 1 - k}!} =
{\pars{-1}^{n} \over \pars{n + 1}!}
\sum_{k = 0}^{n +1}{n + 1 \choose k}\pars{k - 1}^{2}\pars{-1}^{k - 1}
\\[3mm] & =
{\pars{-1}^{n} \over \pars{n + 1}!}\,
\lim_{x\ \to\ -1}\pars{x\,\partiald{}{x}}^{2}\
\sum_{k = 0}^{n +1}{n + 1 \choose k}x^{k - 1}
\\[3mm] & =
{\pars{-1}^{n} \over \pars{n + 1}!}\, \underbrace{%
\lim_{x\ \to\ -1}\pars{x\,\partiald{}{x}}^{2}
\bracks{\pars{1 + x}^{n + 1} \over x}}_{\ds{=\ 0}}
= 0
\end{align}



The RHS $\pars{1}$ vanishes out which show that the sum is independent of $\ds{n}$ and it has the value of $\ds{\ \large\color{#f00}{1}}$.

