How to prove that $n^5 - n$ is a multiple of $5$? Hello I'm new to induction so please bare with me. For this problem I have to use induction to prove: For every integer $n\geq 1$, the number $n^5 − n$ is a multiple of $5$. Can someone please help me figure this out. I've tried to follow the examples in the book, but they were very confusing. 
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$\Large\tt A "Non-Induction" Proof$:

Let $n = p + \delta$ where $p\in{\mathbb N}$. $10|p$.
  $0 \leq \delta \leq 9$. Then,
  $$
n^{5} = \pars{p + \delta}^{5}
= \overbrace{%
p^{5} + 5p^{4}\delta + 10p^{3}\delta^{2} + 10p^{2}\delta^{3} + 5p\delta^{4}}
^{\mbox{Obviosusly, it's a multiple of}\ 10}\ +\ \delta^{5}
$$
  $\delta^{5} \in \braces{\color{#00f}{\large 0},\color{#00f}{\large 1},3\color{#00f}{\large 2},24\color{#00f}{\large 3},102\color{#00f}{\large 4},312\color{#00f}{\large 5},777\color{#00f}{\large 6},1680\color{#00f}{\large 7},3276\color{#00f}{\large 8},5904\color{#00f}{\large 9}}$ such that the last digit of $n^{5}$ is equal to $\delta$. Then, the last digits of $n$ and $n^{5}$ are equal which means that the last digit of $n^{5} - n$ is zero and
  $\color{#00f}{\large 5 | \pars{n^{5} - n}}$. 

A: Clearly for $n = 1$: $1^5 - 1 = $ which is a multiple of $5$ 
Induction Hypothesis $5$ divides $n^5 - n$. We need to show: $5$ divides $(n+1)^5 - (n+1)$.
Now, $(n+1)^5 - (n+1)$ = $n^5 + 5n^4 + 10n^3 + 10n^2 + 5n +1-n -1$. So $5$ divides $(n+1)^5 - (n+1)$.
So by Principle of Mathematical Induction $5$ divides $n^5 - n$ for all $n\in \mathbb{N}$.
A: Base case, $n=1$:  $1^5-1=0$ is divisible by 5.  
Hypothesis: $\exists n \geq 1: 5|(n^5-n)$  
Step:
$$(n+1)^5-(n+1)= (\sum_{k=0}^5\binom{5}{k}n^k) - (n+1) \\ = 1+ 5n+ 10n^2 + 10n^3+5n^4+n^5-n-1=5(n+2n+2n^2+n^4) + n^5-n$$
Now by the hypothesis $5|(n^5-n)$, and clearly $5|[5(n+2n+2n^2+n^4)]$, and therefore:
$$ 5|[(n+1)^5-(n+1)]$$
This completes the proof by induction and it follows that:
$$\forall n \geq 1: 5|(n^5-n)$$
If you don't believe in induction (I don't believe in induction, because it has nothing to do with belief ;-):
$$n^5-n = n(n^4-1) = n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1)$$
Now, if:
$$(n-1 \equiv 0 \bmod 5) \oplus (n \equiv 0 \bmod 5) \oplus (n+1 \equiv 0 \bmod 5)$$ 
We are done. If not, then:
$$(n-2 \equiv 0 \bmod 5) \oplus (n+2 \equiv 0 \bmod 5)$$
Must hold. Now notice that $(n-2)(n+2)=n^2-4 \equiv n^2+1\bmod 5$.    
So in this case, using modular multiplication rules, $n^2+1 \equiv 0 \bmod 5$ must hold.  
So there's no escape! $5|(n^5-n)$.
A: I would like to show another approach without induction method
We know that
$$n^5 - n=n(n^4-1)=n(n^2-1)(n^2+1)=(n-1)n(n+1)(n^2+1)$$
Now Let's focus on another sequence 
$ n>2$ and $k>0$
$$(n-2)(n-1)n(n+1)(n+2)= 5.k$$
Because if we have 5 numbers sequence and always we will have a number in the sequence such form of $5.j$.
Examples:
$3.4.5.6.7$
or 
$4.5.6.7.8$
or
$5.6.7.8.9$
or 
$6.7.8.9.10$
another sequence also must be such form of 
$$(n-1)n(n+1).5= 5.m$$
$$(n-1)n(n+1)(n^2-4)= 5.k$$
To sum up both terms , you can get
$$(n-1)n(n+1)(n^2-4+5)= 5(m+k)$$
$$(n-1)n(n+1)(n^2+1)=n^5 - n= 5.(k+m)$$
A: Proof: 
Base case $n = 1: 1^5 - 1 = 0 = 0*5$ which is a multiple of $5 ( k = 0 )$.
Assume $5$ divides $n^5 - n$. We need to show: $5$ divides $(n+1)^5 - (n+1)$.
$(n+1)^5 - (n+1) = n^5 + 5k + 1 - (n+1) = 5k + n^5 - n$. We have: $5$ divides $5k$, and from induction step: $5$ divides $n^5 - n$. So $5$ divides $(n+1)^5 - (n+1)$.
So: $5$ divides $n^5 - n$ for all $n$.
