Show by induction : $n^7-n$ is a multiple of 7 I have to prove this : "$n^7-n$ is a multiple of 7". This is what I have done this so far :

$P(n):n^7-n$
On putting $n=1,$
$P(1):1^7-1=0$, which is a multiple of 7.
So, $P(1)$ is true.
Let $P(k)$ be a multiple of 7.
So, $k^7-k$ is a multiple of 7.
So, $k^7-k=7m$, where $m$ is any natural number.
Let $P(k+1)$ be true,
Hence, $(k+1)^7-(k+1)$ is a multiple of 7.

Now, how do I evaluate $(k+1)^7$ without using the binomial theorem ? Or is there some other way to do it ?

Without induction:
$$n^7-n=n(n^6-1)=n(n^2-1)(n^4+n^2+1)$$
Now, $$n^4+n^2+1-(n^2-2^2)(n^2-3^2)=14n^2-35\equiv0\pmod7$$
So, $$n^7-n\equiv n(n-1)(n+1)(n-2)(n+2)(n-3)(n+3)\pmod7$$ 
The right hand side is the product of $7$ consecutive integers, hence divisible by $7$
