I am a high school student self-studying number theory and came across this question in the book Challenge and Thrill of Pre-College Mathematics:
Prove that any integer $n$ whose prime factorization is $n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ is a perfect square iff $2|a_i$, and in general $n$ is an $m^{th}$ power iff all $a_i$ is a multiple of $m$.
I managed to prove this for $m=2$, as the question said, but am unable to even get started on the second part, because my proof required an assumption which I do not know if holds for the general case also.
Please check the proof below:
$n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ where $n$ is a perfect square.
Let us assume, if possble, that there exist some powers $a_i\geq 0$ such that $a_i=2b_i+1$.
We know that the number of factors $\tau(n)=(a_1+1)(a_2+1)...(a_k+1)$. We can write each $a_i$ as $a_i= 2b_i$ or $a_i=2b_i+1$.
Now $\tau(n)=(2b_1+1)(2b_2+1)...(2b_i+1+1)...(2b_k+1)$.
$\Rightarrow 2|\tau (n)$
$\color{red}{\textrm{However we know that every perfect square has an odd number of factors.}}$ This contradicts our assumption that $n$ is a perfect square.
Thus, our assumption is wrong and no such $a_i=2b_i+1$ exists.
QED.
Can the assumption highlighted in red be generalized for any number which is an $m^{th}$ power? And if not, is there a better way of doing it than my proof?