3. Find all positive integers $k$ such that the product of the first $k$ primes minus 1 is a positive integer that is a perfect power greater than 1.
3. $k=1$.

Let $2=p_{1}1, k>1$.
Since $a$ is odd and coprime with $p_{i}(1 \leqslant i \leqslant k)$, we know $a>p_{k}$.
Without loss of generality, we can assume $n$ is a prime (if $n=s t$, then we can replace $n$ with $t$ and $a$ with $a^{2}$).
Since $a^{2}+1$ is not a multiple of 3, we have $n>2$.
If there exists $2 \leqslant i \leqslant k, n=p_{i}$, then $p_{i} \mid\left(a^{p_{i}}+1\right)$.
On the other hand, by Fermat's Little Theorem, we have
$$
\begin{array}{l}
a^{p_{i}} \equiv a\left(\bmod p_{i}\right) \Rightarrow p_{i} \mid(a+1) \\
\Rightarrow a \equiv-1\left(\bmod p_{i}\right) \\
\Rightarrow 1-a+a^{2}-\cdots+a^{p_{i}-1} \\
\quad \equiv \underbrace{1+1+\cdots+1}_{p_{i} \uparrow}=p_{i} \equiv 0\left(\bmod p_{i}\right) .
\end{array}
$$

Thus, $p_{1} p_{2} \cdots p_{k}=a^{p_{i}}+1$
$$
=(a+1)\left(1-a+a^{2}-\cdots+a^{p_{i}-1}\right)
$$

is a multiple of $p_{i}^{2}$, which is a contradiction.
Therefore, $a>p_{k}$.
Then $a^{n}+1>p_{k}^{p_{k}}>p_{k}^{k}>p_{1} p_{2} \cdots p_{k}$, which is a contradiction.