If a prime $p$ divides $n^2$ then it also divides $n$ - Is this proof correct?

A shorter way would be to write $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ (where each $\alpha_i>0$) by the Fundamental Theorem of Arithmetic. Then $n^2= p_1^{2\alpha_1}\cdots p_k^{2\alpha_k}$. The only primes that divide $n^2$ are $p_1,\cdots,p_k$ and these clearly also divide $n$.