$1^0 = 1\to 1 =1$
$x^1=x\to x=x\;\forall x$.
$9^2 = 81\to 8+1=9$
$46^5 = 205962976\to 2+0+5+9+6+2+9+7+6=46$
$64^6 = 68719476736\to 6+8+7+1+9+4+7+6+7+3+6 = 64$
$68^7= 6722988818432\to 6+7+2+2+9+8+8+8+1+8+4+3+2 = 68$
$54^8 = 72301961339136\to 7+2+3+0+1+9+6+1+3+3+9+1+3+6=54$
$71^9 = 45848500718449031$
$4+5+8+4+8+5+0+0+7+1+8+4+4+9+0+3+1 = 71$
Given a positive integer $b$, there exists a positive integer $a$ such that the digit sum of $a^b$ is equal to $a$.
Can this be proven? I don't know how to prove it; it was about 3:45am in the morning and I couldn't go to sleep, so I just went on my calculator and messed around because I was bored. That's when I noticed this cool property and decided to share it here.
It is now 4:35am so... I gotta go to bed. I'll see you in some hours and hopefully gather the time to work on this. Sorry about that!
Oh, incidentally, I also noticed that the digit sum of $29^5$ is $23$ and the digit sum of $23^5$ is $29$, so... there are cycles here. Same for $31$ and $34$. Also, the digit sum of $13^2$ is $16$ and the digit sum of $16^2$ is $13$. These cycles seem to only have two numbers involved, but I think regarding the seventh power, there is more than two involved (start with $72^7$ I think), however there is also a cycle between two numbers of seventh powers (between $44$ and $62$). Does this help? I don't know.
I have to go to bed. Good night!
Thank you in advance.