The answer to this question is quite simple (and trivial). Maybe you didn't research the matter properly.
First of all, the problem is "only" a problem for $n \ge 2$ (we don't have to lose our time with merging one square to itself, right?!).
Using induction, we have to prove, at first, that we can divide two squares into a certain number of parts in a way to rearrange those parts into another square.
If the squares are equal to each other, we can do the following (for instance):
On the other side, if the squares are different then:
It's easy to see (and conclude) that the previous method works for any situation (for two arbitrary squares, they can be equal to each other or different; there's no other possibility).
The first induction step is done!
Then, let's assume that we can do our merging with $p$ squares (i.e, your conjecture is valid for $n=p$). If we have $p+1$ squares then, by hypotheses, we can merge the first $p$ squares into one first and then we can do the merging of this one with the $(p+1)$-th square (because we already proved that we can do it with two squares). So, if the merging is possible for $p$ squares, it's also possible for $p+1$.
Then, by induction, your conjecture is proved. $QED$