<s>[INST] $s(n^x)$ is not a perfect square for all $x$ 
Does there exist an $n$ such that $s(n^x)$ is not a perfect square for all positive integers $x$ where $s(m)$ denotes the sum of the digits of a positive integer $m$?

If $n = 5$, for example, then take $x = 8$ to get $5^8 = 390625$, which has digit sum $25 = 5^2$.
If $n = 6$, take $x = 2$ to get $6^2 = 36$, which has digit sum $9 = 3^2$.
If $n = 7$, take $x = 7$ to get $7^7 = 823543$, which has digit sum $25 = 5^2$.
For $n = 19$, take $x = 15$ to get $19^{15} = 15181127029874798299$, which has digit sum $100 = 10^2$.
How can we find such an $n$ or prove that none exists?
 [/INST] Here is a very handwavy argument that none exists.  There are $9 \cdot 10^{k-1}$ numbers with $k$ digits, of which $\sqrt{10^k}-\sqrt{10^{k-1}}=10^{\frac k2}-10^{\frac{k-1}2}=(\sqrt{10}-1)10^{\frac{k-1}2}$ are squares.  We imagine that the numbers of interest are randomly squares or not, so a $k$ digit number has probability $\frac {\sqrt{10}-1}910^{-\frac{k-1}2}$ of being a square.  
$n^m$ has $m \log_{10}n$ digits and the average digit is $4.5$, so the digit sum is about $4.5m\log_{10}n$, which has $\log_{10}(4.5m\log_{10}n)=\log_{10}m+c$ digits for a suitable $c$.  The chance for the digit sum to be square is then $\frac{\sqrt{10}-1}9\cdot 10^{-\frac {\log_{10}m+c}2}=\frac{\sqrt{10}-1}9\cdot10^{-\frac c2}\cdot m^{-\frac 12}$.  The sum of these as $m$ goes from $1$ to $\infty$ diverges, so we expect infinitely many cases where the sum of digits of $n^m$ is a square.  
Of course, someone could find a number where there is a recurring pattern to make the sum of digits nonsquare.  The squares $\bmod 9$ are $0,1,4,7$ so if there were a number that you could prove the digit sums of the powers were always something else we would have an answer.  For example, the digit sums of $10^m$ are always $1$.  In this cases $1$ is a square, so this is not a problem.
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