Saudi Arabian IMO Booklet

Denote the first $n$ primes as $$p_1=2<p_2=3<\cdots <p_n$$ Suppose that $p_1{p}_2\cdots p_n+1=x^k$ 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 $x^k\equiv 1(\mathrm{mod}k)$, which implies $x\equiv 1(\mathrm{mod}k)$, but then by Lifting the Exponent Lemma $$x^k\equiv 1(\mathrm{mod}{k}^2).$$ 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. □