Answer
Let $P(n)$ denote the proposition: "5 divides $n^5 -n$ whenever $n$ is a nonnegative integer."
$P(1)$ is true, since $(1)^5 -(1) = 0$.
Assume that $P(k)$ is true for an arbitrary, nonnegative integer $k$. Then
\begin{align}
(k+1)^5 - (k+1) &= k^5 + 5k^4 + 10k^3 + 10k^2 +5k +1 -k - 1\\
&= [k^5 -k] + [5k^4] + [10k^3] + [10k^2] + [5k]
\end{align}
Each of the bracketed terms is divisible by $2$, so $P(k+1)$ is also true. Therefore by method of mathematical induction $P(n)$ is true - 5 divides $n^5 -n$ whenever $n$ is a nonnegative integer.
Work Step by Step
First, we must give the proposition. We let $P(n)$ denote the proposition: "$5$ divides $n^5 - n$ whenever $n$ is a nonnegative integer."
Next, we must prove our base case $P(1)$ is true by showing that the proposition is true when $n=1$. When $n=1$, $n^5 -n = (1)^5 -(1) =0$. $0$ is divisible by $5$, so therefore $P(1)$ is true.
Finally, we must carry out the inductive step, Assume that $P(k)$ is true; that is, we assume that $k^5 -k$ is divisible by $5$ for an arbitrary, nonnegative integer $k$. Now we must show that this implies that $P(k+1)$ is also true. A useful technique to remember in this proof is the method of binomial expansion using Pascal's triangle.
\begin{align}
(k+1)^5 - (k+1) &= {5 \choose 0}k^5 +{5 \choose 1} k^4 + {5 \choose 2} k^3 + {5 \choose 3} k^2 \\ &+ {5 \choose 4} k^1 + {5 \choose 5} -k - 1 \\
&= k^5 + 5k^4 + 10k^3 + 10k^2 +5k +1 -k - 1\\
&= k^5 - k +5k^4 + 10k^3 +10k^2 + 5k \\
&= [k^5 -k] + [5k^4] + [10k^3] + [10k^2] + [5k]
\end{align}
The first bracketed term is divisible by $5$, via our inductive hypothesis; the rest of the bracketed terms are trivially divisible by $5$. This means that the entire expression must also be divisible by $5$, and that $P(k+1)$ is true. Therefore by method of mathematical induction $P(n)$ is true - $5$ divides $n^5 -n$ whenever $n$ is a nonnegative integer.