10.3. Find all natural $k$ such that the product of the first $k$ prime numbers, decreased by 1, is a perfect power of a natural number (greater than the first).

(V. Senderov)
Answer. $k=1$.

Solution. Let $n \geqslant 2$, and $2=p_{1}<p_{2}<\ldots<p_{k}$; then $k>1$. The number $a$ is odd, so it has an odd prime divisor $q$. Then $q>p_{k}$, otherwise the left side of the equation $(*)$ would be divisible by $q$, which is impossible. Therefore, $a>p_{k}$.

Without loss of generality, we can assume that $n$ is a prime number (if $n=s t$, then we can replace $n$ with $t$, and $a$ with $a^{s}$). Note that $n>2$, since $a^{2}+1$ cannot be divisible by $3=p_{2}$.

We will now show that $n>p_{k}$. Otherwise, we have $n=p_{i}$ for some $1<p_{i} \leqslant p_{k}$. Since $a>p_{k}$ and $n>p_{k}$, it follows that $a^{n}+1>p_{k}^{p_{k}}>p_{1} p_{2} \ldots p_{k}$, which contradicts the equation $(*)$.