Choose $k \equiv -i \mod p_i q_i$ for some distinct primes $p_i, q_i$ by the chinese remainder theorem, $i=1,...,n$. Then $k+i$ is divisible by $p_i q_i$, thus surely not power of a prime. A bit stronger: we also don't allow the numbers to be a nontrivial power of any integer. Then the proof can be adapted to: Choose $k \equiv p_i-i \mod p_i^2$ for some distinct primes $p_i$ by the chinese remainder theorem, $i=1,...,n$. We can assume $k>p_i$ for all such $i$, too. Then $k+i$ is divisible by $p_i$, but not by $p_i^2$. Since also $k+i \neq p_i$, $k+i$ is neither nontrivial power of any integer, nor prime itself.

another proof: these numbers satisfies the condition:$ (2(n + 1))! + 2,(2(n + 1))! + 3,...,(2(n + 1))! + (n + 1)$