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.
 A: 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!
A: $f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)\implies f(n)-nf(n-1)=-(f(n-1)-(n-1)f(n-2))$
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.
A: $f(n)=(n-1)\{f(n-1)+f(n-2)\}$ 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.
