, 2021

For fixed $n$ there are clearly finitely many solutions; we will show that there is no solution with $n > 100$. So, assume $n > 100$. By the AM-GM inequality,
$$\begin{aligned} n! & = 2 n(n-1)(n-2)(n-3) \cdot (3 \cdot 4 \cdots (n-4)) \\ & \leqslant 2(n-1)^{4}\left(\frac{3+\cdots+(n-4)}{n-6}\right)^{n-6} = 2(n-1)^{4}\left(\frac{n-1}{2}\right)^{n-6} < \left(\frac{n-1}{2}\right)^{n-1} \end{aligned}$$
thus $a, b, c < (n-1)/2$.

For every prime $p$ and integer $m \neq 0$, let $\nu_{p}(m)$ denote the $p$-adic valuation of $m$; that is, the greatest non-negative integer $k$ for which $p^{k}$ divides $m$. Legendre's formula states that
$$\nu_{p}(n!) = \sum_{s=1}^{\infty} \left\lfloor \frac{n}{p^{s}} \right\rfloor$$
and a well-known corollary of this formula is that
$$\begin{equation*} \nu_{p}(n!) < \sum_{s=1}^{\infty} \frac{n}{p^{s}} = \frac{n}{p-1} . \tag{$\ominus$} \end{equation*}$$
If $n$ is odd then $a^{n-1}, b^{n-1}, c^{n-1}$ are squares, and by considering them modulo $4$ we conclude that $a, b$ and $c$ must be even. Hence, $2^{n-1} \mid n!$ but that is impossible for odd $n$ because $\nu_{2}(n!) = \nu_{2}((n-1)!) < n-1$ by $(\ominus)$.

From now on we assume that $n$ is even. If all three numbers $a+b, b+c, c+a$ are powers of $2$ then $a, b, c$ have the same parity. If they all are odd, then $n! = a^{n-1} + b^{n-1} + c^{n-1}$ is also odd which is absurd. If all $a, b, c$ are divisible by $4$, this contradicts $\nu_{2}(n!) \leqslant n-1$. If, say, $a$ is not divisible by $4$, then $2a = (a+b) + (a+c) - (b+c)$ is not divisible by $8$, and since all $a+b, b+c, c+a$ are powers of $2$, we get that one of these sums equals $4$, so two of the numbers of $a, b, c$ are equal to $2$. Say, $a = b = 2$, then $c = 2^{r} - 2$ and, since $c \mid n!$, we must have $c \mid a^{n-1} + b^{n-1} = 2^{n}$ implying $r = 2$, and so $c = 2$, which is impossible because $n! \equiv 0 \not\equiv 3 \cdot 2^{n-1} (\bmod 5)$.

So now we assume that the sum of two numbers among $a, b, c$, say $a+b$, is not a power of $2$, so it is divisible by some odd prime $p$. Then $p \leqslant a+b < n$ and so $c^{n-1} = n! - (a^{n-1} + b^{n-1})$ is divisible by $p$. If $p$ divides $a$ and $b$, we get $p^{n-1} \mid n!$, contradicting $(\ominus)$. Next, using $(\ominus)$ and the Lifting the Exponent Lemma we get
$$\begin{equation*} \nu_{p}(1) + \nu_{p}(2) + \cdots + \nu_{p}(n) = \nu_{p}(n!) = \nu_{p}(n! - c^{n-1}) = \nu_{p}(a^{n-1} + b^{n-1}) = \nu_{p}(a+b) + \nu_{p}(n-1) . \tag{$\diamond$} \end{equation*}$$
In view of $(\diamond)$, no number of $1, 2, \ldots, n$ can be divisible by $p$, except $a+b$ and $n-1 > a+b$. On the other hand, $p \mid c$ implies that $p < n/2$ and so there must be at least two such numbers. Hence, there are two multiples of $p$ among $1, 2, \ldots, n$, namely $a+b = p$ and $n-1 = 2p$. But this is another contradiction because $n-1$ is odd. This final contradiction shows that there is no solution of the equation for $n > 100$.