Prove that ${n^5 - n}$ is divisible by 5 I need to prove by induction if ${n^5 - n}$ is divisible by 5 and I have no idea how I would do it. 

I am trying to prove it for several hours now, I started with ${n^5 - n} \mod  5 = 0$ but then I realized that I have no idea how to use modulo for transformations.

The next thing I tried was $(5 + n)\frac{n^5 - n}5 = \frac{(n+1)^5 - n+1}5$ but it didn't get me anywhere.

How would you do it?
Let $\displaystyle f(n): n^5-n$ is divisible by $5$ holds true for $n=m$

$\displaystyle\implies 5\mid(m^5-m)$

Now, $\displaystyle f(m+1)-f(m)=(m+1)^5-(m+1)-(m^5-m)=\sum_{1\le r \le 4}\binom5 rm^r$ which is divisible by $5$ 

as each $\displaystyle\binom5r,1\le r\le4$ is divisible by $5$

Finally, establish the base case i.e., $n=1$

Generalization :

More generally prime $p$ divides $\displaystyle\binom pr$ (can you prove this?) for $1\le r\le p$ or more genrally $p\not\mid r$

So, we can prove  $\displaystyle n^p-n$ will be divisible by $p$ (Fermat's Little Theorem)
I would like to give you a more simple approach to this question:
=>n^5-n is given to you.
So, on taking n common n(n^4-1) is you got.
so this expression has to be divisible by 5 the multiplication of both n and n^4-1 have to divisible by 5. And, Clearly for each number we have n^4-1 have to be divisible by 5.
Now, calling the number
Any number which has even in unit's place like 2,4,6,8
2^4=6
4^4=6
6^6=6
8^6=6
Any number which have even number in unit's place is leading to in 6 and subtracting 1 we got 5 so each expression is divided by 5
And, similarly for odd numbers 
1^4=1
3^4=1
7^4=1
9^4=1
And, subtracting 1 from we got 0 so it is again divisible by 5.
And, when we have 5 at unit's place n is divisible by 5.
Sorry , for the poor presentable form, but I think this will work