Lemma: Let $p$ be a prime. If $p|f(n)$, then $p|n$.

Proof: Suppose $p$ does not divide $n$, then we may write $n = kp + q$, where $k$ is a positive integer and $0 < q < p$.

Since $f(x)$ is a polynomial with integer coefficients, we have
$$kp|f(n) - f(n - kp) \Rightarrow p|f(n) - f(n - kp)$$
Also since $p|f(n)$ and $n = kp + q$, we obtain
$$p|f(n) - f(n - kp) \Rightarrow p|f(q)$$
By the hypothesis we know $f(q)|(q!)^k$, hence $p|(q!)^k$.

But since $0 < p < q$ and $p$ is a prime, $(q!)^k$ cannot be divisible by $p$, a contradiction!

Thus the lemma is proved.

By the lemma we know: for a prime $p$, $f(p)$ has only $p$ as its prime factor, and since $f(p)|(p!)^k$, the value of $f(p)$ can only be one of $1, p, p^2, \dots, p^k$, that is, $k+1$ possibilities.

Consider the values of this function at all primes; since there are infinitely many primes, by the pigeonhole principle there must exist a nonnegative integer $r$ (where $0 \le r \le k$) such that there are infinitely many primes $p$ satisfying $f(p) = p^r$.

Observe that the polynomial $f(x) = x^r$ satisfies: the equation $f(x) - x^r = 0$ has infinitely many roots, hence $f(x) - x^r = 0$ must hold identically, that is, $f(x) = x^r$, where $0 \le r \le k$, gives all solutions of the equation in the problem.