Prove the digital root of a square can be anything other than $2, 3, 5, 6, 8$? The digital root is the sum of the digits, unless that has more than one digit, so then you add up the digits again, until arriving at a single digit, e.g., $28$ -> $2 + 8 = 10$ -> $1 + 0 = 1$.
For what the digital root of a square can be, I just need to give an example of it, right? $0, 1, 4, 9$ are easy enough, and for $7$ I can just do $16$ or $25$ or $169$ or $196$ or $484$ or $529$ etc. (they seem to follow each other like that for some reason).
I'm not a professional mathematician, so please forgive me if what I'm saying is too obvious or too easy. I have looked among the first thousand squares and failed to find a single one with digital root $2, 3, 5, 6$ or $8$, but that doesn't prove anything, because maybe $1001^2$ does something contrary to expectation. I have this feeling that the explanation is in plain sight but for some reason I'm failing to see it.

The digital root is actually equal to the number $\mod 9$ (the remainder when divided by 9). The only difference is that $0$ is replaced by $9$, except of course when the number is $0$.
Looking at the squares $\mod 9$, we see 1, 4, 9, 7, 7, 9, 4, 1, 9, 1, 4 ....
Therefore, this pattern begins to repeat. And we see that the unattainable values are $2, 3, 5, 6, 8$.
