Prove n is a perfect square if and only if $n^7$ is a perfect square, n $\in$ natural numbers

Question

Proving one way is very simple

If n is a perfect square then

n = $a^2$

$n^7$ = $a^{14}$

$n^7$ = $a^7$($a^7$) which is obviously a perfect square for some integer a.

It's the if $n^7$ is a perfect square then n is a perfect square part that is giving me trouble. I have already tried to get hints from others and the one hint I was given was to write $n^7$ as a product of primes, but I have no clue where to go from there. I believe it can be done using modulus but I honestly have no idea.

Any hints would help thanks.

Answer 1

Maybe it is helpful to observe that $n^6$ is a perfect square and $n=\frac{n^7}{n^6}$.