Is square removal easier than factoring?
We can show that if all $\alpha_i$ are different, then square removal and factoring of $n$ are equally hard.

It is obvious, that if we can factor $n$, we can also compute square removal of $n$.

The other direction is a bit more tricky. First compute the square removal of $n$ and let's call this $m$. From the definition it follows that $m$ divides $n$. Divide by $m$ repeatedly until we reach a number, which is not divisible by $m$, I'll call this $x$

$$x = \frac{n}{m^b}$$

Now compute $p = \frac{m}{\gcd(x, m)}$, if $p$ had more than one prime factor, those would have identical $\alpha_i$ in the original product $n$, which contradicts the assumption about all $\alpha_i$ being different, hence $p$ is a prime factor in $n$.

Knowing one prime factor in $n$ it is possible to reduce the problem to factoring a smaller $n'$, which satisfy the same criteria, so the algorithm can be repeated.

We can also show that if all $\alpha_i$ are equal, then square removal is easy. That's because $\alpha_i$ can only have logarithmic size in $n$, and every possible $\alpha_i$ can be tested by computing the $\alpha_i$th root of $n$.

However the result obtained that way will be incorrect if the $\alpha_i$ were not equal. The obtained result will be correct iff it is square free. And @joro pointed out, that no polynomial algorithm is known to decide if a number is square free.

So for some $n$ square removal and factoring is equivalent. For other $n$ square removal is easy. Telling those two cases apart appears to be hard.