12・138 Find the natural number solutions of the equation $(y+1)^{x}-1=y$!
(Bulgarian Mathematical Olympiad, 1982)
［Solution］Let natural numbers $x$ and $y$ satisfy the equation.
First, we prove that $p=y+1$ is a prime number.
If $p$ is composite, then a factor $d$ of $p$ satisfies $1<d<y+1$, so $d \mid y!$, and $d \mid (y+1)^{x}$, thus
$$
1=(y+1)^{x}-y! \equiv 0 \quad(\bmod d),
$$

which is impossible.
Therefore, if the equation has natural number solutions $x, y$, then $y+1$ is a prime number, and thus the prime number $p$ satisfies the equation
$$
p^{x}-1=(p-1)!.
$$

If $p \geqslant 7$, then dividing both sides of the equation by $p-1$ gives
$$
p^{x-1}+p^{x-2}+\cdots+1=(p-2)!,
$$

and when $p \geqslant 7$,
$$
2<\frac{p-1}{2} \leqslant p-2 \text{, and } \frac{p-1}{2} \in \mathbb{N}.
$$

So the right-hand side of the equation can be divided by $2 \cdot \frac{p-1}{2}=p-1$.
Since $p$ is a prime number, then
$$
\begin{array}{c}
p^{i} \equiv 1^{i} \quad(\bmod (p-1)), \quad i=0,1, \cdots, x-1. \\
\sum_{i=0}^{x-1} p^{i}=(p-2)! \equiv \sum_{i=0}^{x-1} 1^{i}=x \quad(\bmod (p-1)),
\end{array}
$$

which means
$$
x \equiv 0 \quad(\bmod (p-1)).
$$

Thus,
$$
x \geqslant p-1.
$$

However, $p^{x-1}<p^{x-1}+\cdots+1=(p-2)!<(p-2)^{p-2}<p^{p-2}$,
so
$$
\begin{array}{c}
x-1<p-2, \\
x<p-1,
\end{array}
$$

which contradicts $x \geqslant p-1$.
Therefore, it must be that
$$
p<7.
$$

Thus, $p$ can only be $2,3,5$.
If $p=2$, then
$$
2^{x}-1=1,
$$
so
$$
x=1, y=1.
$$

If $p=3$, then
$$
3^{x}-1=2,
$$
so
$$
x=1, y=2.
$$

If $p=5$, then
$$
5^{x}-1=24,
$$
so
$$
x=2, y=4.
$$

Therefore, the original equation has three sets of natural number solutions $(x, y)$:
$$
(x, y)=(1,1),(1,2),(2,4).
$$