<s>[INST] $n^5-n$ 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!
 [/INST] $$n^5-n=n(n^2+1)(n+1)(n-1)= n(n^2-4)(n+1)(n-1)+5n(n-1)(n+1)=(n-2)(n-1)n(n+1)(n+2)+5n(n-1)(n+1)$$
$(n-2)(n-1)n(n+1)(n+2)$ is even and divisible by 5, since it is the product of 5 consecutive integers.
$5(n-1)n(n+1)$ is also even and divisible by $5$.
Note: Both expressions are also divisible by $3$, so $n^5-n$ is actually divisible by $30$!
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