Show that for any prime $p$ and any integer $m$, $m^p + (p − 1)! m$ is divisible by $p$. I'm currently working on the following exercise:

Show that for every prime number $p$ and every integer $m$, the number
  $m^p + (p − 1)! m$ is divisible by $p$.

What I'm doing is the following:
$$m(m^{p-1}+(p-1)!)\equiv 0\mod p$$
$$m((1)+(-1))\equiv 0\mod p$$
$$m(0)\equiv 0 \mod p$$
$$0 \equiv 0 \mod p.$$
Is a valid proof? Am I missing something? Any hint or help will be really appreciated.
 A: 
Show that for every prime number $p$ and every integer $m$, the number
  $mp+(p−1)!m$ is divisible by $p$.

This can be proved via two theorems:  Fermat's Little Theorem and Wilson's Theorem.
$\,m^p+(p-1)!m \;=\; m^p-m+(p-1)!m+m \;=\; m^p-m+m((p-1)!+1)$
By Fermat's little theorem, $\,m^p-m\,$ is an integer multiple of $p$, since $p$ is prime.
By Wilson's Theorem, $\, (p-1)!+1\,$ is an integer multiple of $\,p\,$, since $\,p\,$ is prime.
So, for some $\, k,\,l \in \mathbb{Z}\text{,}\,$ we have
$$m^p-m+m((p-1)!+1) \;=\; kp + m(lp) \;=\; p(k+ml)\text{.}$$
Thus $\,m^p+(p−1)!m\,$ is divisible by $p$.

I believe the error in your proof came from assuming that $\,m^{p-1}\equiv1\,(\text{mod}\; p)\text{.}\,$ This is only true if we assume $m$ is not divisible by $p$.
A: Let $p$ be a prime number and $m$ an integer.  
By Fermat's little theorem, $m^p \equiv m \pmod p$, and by Wilson's theorem $(p-1)! 
\equiv -1 \pmod p$.
Therefore, $ m^p + (p-1)!m \equiv m+(-1) \times m \equiv m-m \equiv 0 \pmod p$,
so $p$ divides $m^p+(p-1)!m.$
