Ukrainian National Mathematical Olympiad

From the problem statement, it is obvious that $k > m$ and $k > n$. We rewrite the given equation as
$$m n ((m-1)! + 1)((n-1)! + 1) = k ((k-1)! + 1).$$

Since $(k-1)!$ is divisible by $m$ and $n$, $(k-1)! + 1$ is not divisible by any factor of these numbers. Hence, $k \nmid m n$. Suppose $n \ge m$. If $m \ge 2$, then
$$\begin{aligned} 4(n!)^2 &\ge 4 m! n! = (2 m!) (2 n!) \ge (m! + m)(n! + n) = k! + k > (m n)! \ge (2 n)! \\ 4 n! n! &> (2 n) \cdot (2 n - 1) \cdots (n + 1) \cdot n! \Rightarrow 4 n! > (2 n) \cdot (2 n - 1) \cdots (n + 1) \\ 4 n \cdot (n - 1) \cdot (n - 2) \cdots 2 \cdot 1 &> (2 n) \cdot (2 n - 1) \cdots (2 n - 2) \cdots (n + 2) \cdot (n + 1) \\ 4 &> \frac{2 n}{n} \cdot \frac{2 n - 1}{n - 1} \cdots \frac{n + 1}{1} \Rightarrow \text{contradictions for } n \ge 2. \end{aligned}$$
Let us show that for $n \ge 2$ this yields a contradiction. On the right-hand side, each factor except the first and last one is greater than the ones on the left. Now, let us show that the product of the first and last factors is still greater on the right-hand side: $4 n \cdot 1 > (2 n) \cdot (n + 1) > 6 n$.

Therefore, there is only one option $m = 1$, hence,
$$2 \cdot n! + 2 n = k! + k \ge (n + 1)! + (n + 1) > (n + 1)! + n \Rightarrow n > n! (n - 1) \Rightarrow n = 1 \Rightarrow m = 1.$$

Then, we easily get $k = 2$.