Prove that the equation $u^a - u^b = n!$ has finitely many solutions $(a,b,n)$ in positive integers. 
(China TST 2004 Day 1). Let $u$ be a fixed integer. Prove that the equation $u^a - u^b = n!$ has finitely many solutions $(a,b,n)$ in positive integers. Here is the original source of the problem.


Note: it's possible one has to additionally assume that $u\neq -1$. If it turns out that the claim is false for some $u\neq -1$, then just assume $u$ must be positive for convenience.

For a positive integer k and prime p, let $v_p(k)$ denote the highest exponent of p dividing k. If $u=0,1$, there are no solutions so assume $u\neq 0,1$. If $u=-1$, then $u^a - u^b$ can only take on the values $0, 2,$ and $-2$. But $2$ can be obtained infinitely many times, namely whenever a is even and b is odd, so it seems we need to assume $u\neq -1$. First consider the case when $u > 1$. Then $a > b$ clearly, so $u^b$ must divide $n!$. Also, $u^{a-b}-1 \ge 1$ is coprime to $u$ and divides $n!$. Equality only holds when $u=2$ and $a=b+1$. But in that case, we have $2^b = n!$, which only holds for $b=1$ as $n!$ is divisible by $3$ for $n\ge 1$. Since we're only interested in showing the given equation has finitely many solutions, we may as well assume $u^{a-b}-1 > 1$ so that it has a prime factor. I know that for any prime p, $v_p(n!) = \sum_{i\ge 1}\lfloor n/p^i\rfloor$. In particular, if we let $p$ be a prime dividing $u$ with $v_p(u) = k$, then from above, $v_p(n!) = kb$.  $n!$ grows faster than $u^n$ as $n\to\infty$ for any integer u. Clearly we cannot have $a = b,$ so it could be useful to consider when $a<b$ and when $a > b$ separately. It could be useful to use inequalities and show that when either $a$ or $b$ or $n$ exceed a certain bound, there are no more solutions. I know a fairly nontrivial fact that the last nonzero digit of $n!$ actually forms a nonperiodic sequence, though that's likely not useful for this problem. I'm also unsure whether it'll be useful to consider orders of powers of primes.
 A: You've made a good start, including suggesting possibly using the orders of the powers of primes. Actually, the solution below uses the multiplicative order of a prime.
As Sil's comment indicates, the AoPS thread Chinese factorial equation references the same problem, with that thread providing several solutions having varying degrees of validity. I'm adapting (including making a few minor corrections) and expanding on the approach used in Post #$12$.
For any $u \gt 1$, rewrite the equation as
$$n! = u^b(u^m - 1) \tag{1}\label{eq1A}$$
where $m = a - b \gt 0$. Next, note there's at most one solution for any fixed $n$ since, for each prime factor $q$ of $u$, we require $b = \frac{\nu_q(n!)}{\nu_q(u)}$. If just one integral value of $b$ works in all cases, then $a$ would be uniquely determined from \eqref{eq1A} based on $b$ and $n$, with there then being just one solution if $a$ is an integer. Thus, it's sufficient to show that only finitely many $n$ are solutions, with this answer doing this by proving that $n$ is bounded above.
Consider any prime $p$ which doesn't divide $u$, with $d$ being the multiplicative order of $u$ modulo $p$, i.e., $d = \operatorname{ord}_p(u)$. For any $n$ at least as large as the smallest prime factor of $u$ (so $b \gt 0$), and also with $n \ge p$, we have $p \mid n! \; \to \; p \mid u^m - 1$, so $d \mid m$. Since $u^m = \left(u^{d}\right)^{m/d}$, then by the Lifting-the-exponent lemma,
$$\nu_p(u^m-1) = \nu_p(u^d-1) + \nu_p\left(\frac{m}{d}\right) \le \nu_p(u^d-1) + \log_p\left(\frac{m}{d}\right) \tag{2}\label{eq2A}$$
With the constant
$$c_1 = \nu_p(u^d-1) - \log_p(d) \tag{3}\label{eq3A}$$
as well as using $\log_p\left(\frac{m}{d}\right) = \log_p(m) - \log_p(d)$, we get from \eqref{eq2A} that
$$\nu_p(u^m-1) \le \log_p(m) + c_1 \tag{4}\label{eq4A}$$
Next, Legendre's formula gives that
$$\nu_p(u^m-1) = \nu_p(n!) = \sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^{i}}\right\rfloor \gt \frac{n}{p} - 1 \tag{5}\label{eq5A}$$
Combining \eqref{eq4A} and \eqref{eq5A} results in
$$\log_p(m) + c_1 \gt \frac{n}{p} - 1 \; \; \; \to \; \; \; m \gt p^{\frac{n}{p} - c_2} \tag{6}\label{eq6A}$$
where $c_2 = c_1 + 1$. We then get
$$n^n \gt n! = u^b(u^m - 1) \gt u^m \gt u^{p^{\frac{n}{p} - c_2}} \tag{7}\label{eq7A}$$
Taking the natural logarithm of each side gives
$$n\ln(n) \gt p^{\frac{n}{p} - c_2}\ln(u) \tag{8}\label{eq8A}$$
However, this can be true for at most a finite number of $n$ since the left side is less than a polynomial of $n$ (e.g., $n^2$), but the right side is a growing exponential function in $n$.
