Sindarov V.A.

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

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Let $n \geq 2$, and $2=p_{1}1$; 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 not the case. 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 show that $n>p_{k}$. Indeed, otherwise $n=p_{i}$, where $i \leq k$. Then $a^{p_{i}}+1$ is divisible by $p_{i}$; on the other hand, by Fermat's Little Theorem, $a^{p_{i}}-a$ is divisible by $p_{i}$. Since $a^{p_{i}}+1=(a+1)\left(a^{p_{i}-1}-a^{p_{i}-2}+\ldots-a+1\right)$, and $a+1=\left(a^{p_{i}}+1\right)-\left(a^{p_{i}}-a\right)$ is divisible by $p_{i}$ and $a^{p_{i}-1}-a^{p_{i}-2}+\ldots-a+1 \equiv 1+1+\ldots+1=p_{i} \equiv 0\left(\bmod p_{i}\right)$, then $a^{p_{i}}+1$ is divisible by $p_{i}^{2}$, which contradicts the condition.

Thus, $a>p_{k}$ and $n>p_{k}$, from which $a^{n}+1>p_{k}^{p_{k}}>p_{1} p_{2} \ldots p_{k}$, which contradicts the equation $(*)$.

## Answer

$k=1$.