5. Let $p_{1}, p_{2}, \ldots, p_{k}$ be distinct prime numbers. Determine the number of natural numbers of the form $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}, \alpha_{i} \in \mathbb{N}$ for which

$$
\alpha_{1} \alpha_{2} \ldots \alpha_{k}=p_{1} p_{2} \ldots p_{k}
$$
Solution. Clearly, from $p_{1} p_{2} \ldots p_{k}=\alpha_{1} \alpha_{2} \ldots \alpha_{k}$ it follows that for every $i \in\{1, \ldots, k\}$, $p_{i} \mid \alpha_{s}$ for some $s \in\{1,2, \ldots, k\}$. Given this, we will construct an ordered $k$-tuple $(\alpha_{1}, \ldots, \alpha_{k})$ of natural numbers for which the equality $p_{1} p_{2} \ldots p_{k}=\alpha_{1} \alpha_{2} \ldots \alpha_{k}$ holds. We take the $k$-tuple $(1,1, \ldots, 1)$. We multiply one (arbitrary) element of this $k$-tuple by the number $p_{1}$, which means that for $p_{1}$ we have $k$ possibilities. Further, in the resulting $k$-tuple, we multiply one (arbitrary) element by $p_{2}$, which means that for $p_{2}$ we have $k$ possibilities. Continuing this process, we find that for each of the prime numbers $p_{3}, \ldots, p_{k}$ we have $k$ possibilities. Therefore, we obtain that there are $k^{k}$ ordered $k$-tuples $\left(\alpha_{1}, \ldots, \alpha_{k}\right)$ for which $p_{1} p_{2} \ldots p_{k}=\alpha_{1} \alpha_{2} \ldots \alpha_{k}$. Clearly, each of these $k$-tuples defines a number of the form $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}, \alpha_{i} \in \mathbb{N}$, and different ordered $k$-tuples define different numbers of the desired form. Indeed, if $\left(\alpha_{1}, \ldots, \alpha_{k}\right) \neq\left(\beta_{1}, \ldots, \beta_{k}\right)$, then $\alpha_{i} \neq \beta_{i}$ for at least one $i \in\{1,2, \ldots, k\}$, so $p_{i}^{\alpha_{i}} \neq p_{i}^{\beta_{i}}$, from which it follows that

$$
p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}} \neq p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \ldots p_{k}^{\beta_{k}}
$$

Conversely, every number of the form $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}, \alpha_{i} \in \mathbb{N}$ for which $\alpha_{1} \alpha_{2} \ldots \alpha_{k}=p_{1} p_{2} \ldots p_{k}$ defines an ordered $k$-tuple $(\alpha_{1}, \ldots, \alpha_{k})$ of natural numbers for which the equality $p_{1} p_{2} \ldots p_{k}=\alpha_{1} \alpha_{2} \ldots \alpha_{k}$ holds.

Finally, the number of natural numbers of the form $p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}, \alpha_{i} \in \mathbb{N}$ for which $\alpha_{1} \alpha_{2} \ldots \alpha_{k}=p_{1} p_{2} \ldots p_{k}$ is equal to $k^{k}$.