Denote the first $n$ primes as
$$p_1 = 2 < p_2 = 3 < \dots < p_n$$
Suppose that $p_1p_2\cdots p_n + 1 = x^k$ for some integers $x, k \ge 2$. We can assume WLOG that $k$ is prime since $x^{kt} = (x^t)^k$. Obviously, $x$ has no prime factors not exceeding $p_n$, so $x > p_n$ and consequently $k < n < p_n$ is one of the first $n$ primes. Now $x^k \equiv 1 \pmod{k}$, which implies $x \equiv 1 \pmod{k}$, but then by Lifting the Exponent Lemma
$$x^k \equiv 1 \pmod{k^2}.$$
This is a contradiction, as $x^k - 1 = p_1 \cdots p_n$ is not divisible by the square of any prime. Thus, there does not exist any $k$ satisfying the problem. □