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 ?
 A: You don't have to fully carry out the binomial theorem, if you know that $7$ divides each $\binom{7}{k}$ except for the outer ones. That is:
$$(k+1)^7=k^7+7(\cdots)+1$$
To prove that $7$ divides each $\binom{7}{k}$ except for the outer ones, consider the formula $\binom{7}{k}=\frac{7!}{k!\,(7-k)!}$, and how $7$ is a factor in the numerator but could not be cancelled out in the denominator .
A: I would suggest the following method over induction

If $n$ and $7$ are coprime then by little fermat theorem $n^6-1\equiv 0 \mod 7$.
Else $n\equiv 0 \mod 7$
Hence $n\cdot\left(n^6-1\right)=n^7-n\equiv 0 \mod 7$

And for an inductive proof $$\left(k+1\right)^7-k-1=k^7-k+7\cdot\left(k^6+3k^5+5k^4+5k^3+3k^2+k\right)\equiv0 \mod 7$$
As $k^7-k$ and the remaining term are both divisible by $7$
A: Without induction: $$n^7-n \;=\; (n-3)(n-2)(n-1)n(n+1)(n+2)(n+3) + 7(2n^5-7n^3+5n)$$ First term on the RHS is a product of $7$ consecutive numbers, one of which must be a multiple of $7$.
A: 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 $7$
