$n^5-n$ is divisible by $10$? I was trying to prove this, and I realized that this is essentially a statement that $n^5$ has the same last digit as $n$, and to prove this it is sufficient to calculate $n^5$ for $0-9$ and see that the respective last digits match. Another approach I tried is this: I factored $n^5-n$ to $n(n^2+1)(n+1)(n-1)$. If $n$ is even, a factor of $2$ is guaranteed by the factor $n$. If $n$ is odd, the factor of $2$ is guaranteed by $(n^2+1)$. The factor of $5$ is guaranteed if the last digit of $n$ is $1, 4, 5, 6,$ $or$ $9$ by the factors $n(n+1)(n-1)$, so I only have to check for $n$ ending in digits $0, 2, 3, 7,$ $and$ $8$.  However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks!

Clearly, $n$ and $n^5$ are of the same parity. Hence, $2 \vert (n^5-n)$.
To check for divisibility by $5$, note that
\begin{align}
n^5 - n & = (n^2+1)n(n+1)(n-1)\\
& = (n^2-4+5)n(n+1)(n-1)\\
& = (n^2-4)n(n+1)(n-1) + 5n(n+1)(n-1)\\
& = (n-2)(n-1)n (n+1)(n+2) + 5n(n+1)(n-1)\\
\end{align}
Clearly, $5 \vert 5n(n+1)(n-1)$. Also, $(n-2)(n-1)n (n+1)(n+2)$ is a product of $5$ consecutive numbers and hence $5$ divides it. 
