How to prove $n^5 - n$ is divisible by 30 without reduction

By the Fermat little theorem, $n^{5} \equiv n \mod 5$ i.e. $n^{5} - n \equiv 0 \mod 5$.

I'm going to take a leap, and suppose that you are doing exercises from the first section of Ireland and Rosen. In that case, I think that FLT is not the 'anticipated' method of solution.

So let's start from your factorization, $(n-1)n(n+1)(n^2 + 1)$

If $n$ is of the form $5k$, $5k-1$, or $5k+1$, we're done from the first three factors. What if $n$ is of the form $5k \pm 2$?

I claim to you that $n^2 + 1$ is always divisible by $5$ if $n = 5k \pm 2$. Perhaps the easiest way to see this is through strict computation. Or you could note that the constant term is either $5$ or $10$. In any case, I leave that part to you: can you show that $n^2 + 1$ is divisible by $5$ if $n = 5k \pm 2$?

Using combinatorial polynomials: $$ \begin{align} n^5-n &=120\binom{n}{5}+240\binom{n}{4}+150\binom{n}{3}+30\binom{n}{2}\\ &=30\left(4\binom{n}{5}+8\binom{n}{4}+5\binom{n}{3}+\binom{n}{2}\right) \end{align} $$ Note: The combinatorial polynomial expansion for a known polynomial, $P(n)$, is not as hard as it might seem. The coefficient, $a_0$, of $\binom{n}{0}$ is simply $P(0)$. If the coefficients, $a_j$, for $\binom{n}{j}$ are known for $j<k$, then the coefficient for $\binom{n}{k}$ is $$ P(k)-\sum_{j=0}^{k-1}a_j\binom{k}{j} $$