Sum of digits of square number raised to itself From testing a few different square numbers, it seems to be the case that when raising a square number to the power of itself, the sum of the digits of the result satisfy the property that the sum of their digits is the square number itself.
I realise the above sentence is quite wordy, so as an example, consider $4^4$. We know that $4^4=256$ and that $2+5+6=13$. It is also the case that $1+3=4$, i.e., the square number itself.
Is this true for any square number? And if so, how can one prove it?
 A: Your original claim is poorly worded in that it does not adequately distinguish between sums of digits and sums of sums of digits and what numbers should be compared against one another.  As written the claim is false, as pointed out by other users, since you have the sum of digits or the sum of the sum of the digits not equaling the original number itself in some cases as is the case for example for $25^{25}$ not having sum of digits or sum of sum of digits equal to $25$.
If you were to instead talk about the repeated sum of digits for both the original number and the number to the power of itself, then we do actually have a true statement.

Claim: for perfect square $x=n^2$ one has that $x^x\equiv x\pmod{9}$

Proof by cases:
As $x=n^2$ it follows that $x$ is equivalent to one of $0,1,4,$ or $7$ modulo $9$
In the first case, we have $x^x\equiv 0^x\equiv 0\pmod{9}$ trivially.
Similarly in the second case we have $x^x\equiv 1^x\equiv 1\pmod{9}$
In the third case, $x^x\equiv 4^x\equiv 4^{9k+4}\equiv (4^{3})^{3k}\cdot 4^3\cdot 4\equiv 1^{3k}\cdot 1\cdot 4\equiv 1\pmod{9}$ noting that $4^3=64=9\cdot 7 + 1$
Finally, for the fourth case we have $x^x\equiv 7^{9k+7}\equiv (7^3)^{3k}\cdot 7^3\cdot 7\equiv 1^{3k}\cdot 1\cdot 7\equiv 7\pmod{9}$ just like in the previous case.
