Show that $\frac{f(n)}{n!}=\sum_{k=0}^n \frac{(-1)^k}{k!}$ 
Consider a function $f$ on non-negative integer such that $f(0)=1,f(1)=0$ and $f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)$ for $n \geq 2$. Show that 
  $$\frac{f(n)}{n!}=\sum_{k=0}^n \frac{(-1)^k}{k!}$$

Here
$$f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)$$
$$\implies f(n)=(n-1)(f(n-1)+f(n-2))$$
Then I am stuck.

$f(n)=(n-1)\{f(n-1)+f(n-2)\}$ for $n\geq 2$
Let for $n\in\mathbb N,~S_n:\frac{f(n)}{n!}=\sum_{k=0}^n \frac{(-1)^k}{k!}.$
It's a matter of simple verification that $S_0,~S_1$ are true. 
Let $S_n$ be true $\forall~n\leq m~(\in\mathbb N).$ Then, 
$\dfrac{f(m+1)}{(m+1)!}$
$=\dfrac{m\{f(m)+f(m-1)\}}{(m+1)!}$
$=\dfrac{m}{m+1}.\dfrac{f(m)}{m!}+\dfrac{1}{m+1}.\dfrac{f(m-1)}{(m-1)!}$
$=\dfrac{m}{m+1}\sum_{k=0}^m\dfrac{(-1)^k}{k!}+\dfrac{1}{m+1}\sum_{k=0}^{m-1}\dfrac{(-1)^k}{k!}$
$=\dfrac{m+1-1}{m+1}\sum_{k=0}^m\dfrac{(-1)^k}{k!}+\dfrac{1}{m+1}\sum_{k=0}^{m-1}\dfrac{(-1)^k}{k!}$
$=\sum_{k=0}^m\dfrac{(-1)^k}{k!}-\dfrac{1}{m+1}\{\sum_{k=0}^m\dfrac{(-1)^k}{k!}-\sum_{k=0}^{m-1}\dfrac{(-1)^k}{k!}\}$
$=\sum_{k=0}^m\dfrac{(-1)^k}{k!}-\dfrac{1}{m+1}.\dfrac{(-1)^m}{m!}$
$=\sum_{k=0}^m\dfrac{(-1)^k}{k!}+\dfrac{(-1)^{m+1}}{(m+1)!}$
$=\sum_{k=0}^{m+1}\dfrac{(-1)^k}{k!}$
$\implies S_{m+1}$ is true. Hence by the principle of mathematical induction the result follows.
