I did some paper-and-pencil work with a help from Wolfram Alpha and discovered something that I cannot prove due to a lack of inspiration but it seems to be true (judging from the few cases I calculated).
Denote by $RDS_{10}(a)$ the reduced digit sum of number $a$ in base $10$. By the reduced digit sum I mean a sum of digits applied so many times until we arrive at a one-digit number, so, for example we have $RDS_{10}(17)=1+7=8$ and $RDS_{10}(999)=9+9+9=27=2+7=9$.
Let us take a closer look at numbers for which we have $RDS_{10}(a)=7$ and look at $RDS_{10}(a^4)$ of those numbers.
So we have:
$RDS_{10}(7)=7$ and $RDS_{10}(7^4)=2401=2+4+0+1=7$
$RDS_{10}(16)=1+6=7$ and $RDS_{10}(16^4)=65536=6+5+5+3+6=25=2+5=7$
$RDS_{10}(25)=7$ and $RDS_{10}(25^4)=7$
$RDS_{10}(34)=7$ and $RDS_{10}(34^4)=7$
and so on and so on...
Of course that I checked some other cases but I do not want that this question becomes too lenghty by listing verified cases (I must admit that I did not check too much of them, maybe just 10-15).
Is this always true:
If $RDS_{10}(a)=7$ then $RDS_{10}(a^4)=7.$
The reduced digitsum of $N$ is just $N$ modulo $9$, unless $N$ is divisible by $9$. Because of $$7^4\equiv 7\mod 9$$ your claim is true.