3.10. (НРБ, 82; Australia, 83). Solve the equation

$$
(y+1)^{x}-1=y!
$$

in natural numbers.
3.10. Let the pair of numbers $x, y \in \mathbf{N}$ satisfy the equation. Note that the number $p=y+1$ is prime. Indeed, if there exists a divisor $d$ of the number $p$ satisfying the condition $1<d<y+1$, then we have $y+1 \vdots d$ and

$$
1=(y+1)^{x}-y!\equiv 0(\bmod d)
$$

which is impossible. Therefore, we have the equation

$$
p^{x}-1=(p-1)!
$$

from which it follows that $p<7$. Indeed, suppose that $p \geqslant 7$. Then, by canceling both parts of the equation by $p-1$, we get the equality

$$
p^{x-1}+\ldots+1=(p-2)!
$$

in which the right-hand side is divisible by $2 \cdot(p-1) / 2=p-1$ (since $2<(p-1) / 2 \leq p-2$ for $p \geqslant 7$ and $(p-1) / 2 \in \mathbb{Z})$, while the left-hand side is the sum of $x$ numbers of the form

$$
p^{i} \equiv 1^{1}=1(\bmod (p-1)), i=0,1, \ldots, x-1
$$

Therefore, $x \equiv 0(\bmod (p-1))$, which means $x \geqslant p-1$. But this is impossible, since

$$
p^{x-1}<p^{x-1}+\ldots+1=(p-2)!<(p-2)^{p-2}<p^{p-2}
$$

i.e., $x<p-1$. Thus, the prime number $p<7$ can only take the values 2, 3, or 5. If $p=2$, then

$$
2^{x}-1=1 \quad \text{and} \quad x=1, y=1
$$

If $p=3$, then

$$
3^{x}-1=2 \text{ and } x=1, y=2
$$

If $\rho=5$, then

$$
5^{x}-1=24 \text{ and } x=2, y=4
$$

Thus, we obtain that the original equation is satisfied by only three pairs of natural numbers $(x ; y):(1 ; 1),(1 ; 2)$ and $(2 ; 4)$.