$(nm)$th prime is less than the $m$th power of $n$th prime

Question

Let $n, m >1$. Prove that $P_{n\cdot m}<P_{n}^m$, where $P_{n\cdot m}$ is the $n\cdot m$th prime and $P_{n}^m$ is the nth prime of degree $m$.

Answer 1

Hint:

There's always a prime between $n$ and $2\cdot n$.