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.

Let
$$ g(n) = \sum_{k=0}^n(-1)^k\frac{n!}{k!} \tag{1} $$
then
\begin{align} 
g(n) &= \sum_{k=0}^n(-1)^k\frac{n!}{k!} \\
&= n\sum_{k=0}^{n-1}(-1)^k\frac{(n-1)!}{k!} + (-1)^n\frac{n!}{n!} \\ \\ \\
&= ng(n-1)+(-1)^n \\ \\
&= (n-1)g(n-1) +g(n-1)+(-1)^n \\
&= (n-1)g(n-1)+\Big((n-1)g(f-2)+(-1)^{n-1}\Big) + (-1)^n\\
&= (n-1)g(n-1) + (n-1)g(n-2)
\end{align}
so for any function $g$ that fulfills $(1)$ we have that
$$ g(n)+g(n-1)=ng(n-1)+(n-1)g(n-2) $$
and with $g(0) = 1 = f(0)$, $g(1) = 0 = f(1)$ we conclude $g \equiv f$.
Cheers!
