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.

$f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)\implies f(n)-nf(n-1)=-(f(n-1)-(n-1)f(n-2))$
Let $D(n)=f(n)-nf(n-1)$ then, previous relation becomes,
$D(n)=-D(n-1)=(-1)^2D(n-2)=\cdots=(-1)^{n-1}D(1)=(-1)^n$ (because $D(1)=f(1)-f(0)=-1$)
Thus, $$D(n)=(-1)^n\implies f(n)=nf(n-1)+(-1)^n=n!(\frac{f(n-1)}{(n-1)!}+\frac{(-1)^n}{n!})$$
$$=n((n-1)f(n-2)+(-1)^{n-1})+(-1)^n=n!(\frac{f(n-2)}{(n-2)!}+\frac{(-1)^{n-1}}{(n-1)!}+\frac{(-1)^n}{n!})$$
Continuing in this fashion gives 
$$f(n)=n!(\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^n}{n!})$$
$\implies \frac{f(n)}{n!}=\sum_{k=0}^n\frac{(-1)^k}{k!}$
Hence, proved.
