Are all powers of 5 Friedman numbers?

Question

Powers of 5 seem to have a quite interesting property. Not only do the all seem to be Friedman numbers in base 10, it also seems that they don't require digit concatenation and they their 'Friedman expressions' can always be written in the form $5^\text{something}$. Here's some examples:

$$ \begin{align} 5^2 = 25 =& 5^{2}\\ 5^3 = 125 =& 5^{1 + 2}\\ 5^4 = 625 =& 5^{6 - 2}\\ 5^5 = 3125 =& 5^{1 \times 3 + 2}\\ 5^6 = 15625 =& 5^{6^2 / (1 + 5)}\\ \vdots \\ 5^{14} = 6103515625 =& 5^{6 + 1 + 3 + 5 - 1 + 0 \times 5 \times 6 \times 2}\\ 5^{15} = 30517578125 =& 5^{3\times 5 + 0 \times \cdots}\\ \vdots \end{align} $$

The problem is though, I haven't proved it for all $n$. Intuitively, it seems like it should be true: it's true for small $n$, and for large $n$, $5^n$ has so many digits that making $n$ should be easy (and then some of the remaining digits can make zero so all the remaining digits can be ignored). But I have no clue how to turn this into an actual mathematical argument.

(It's slightly out of the scope of this question, but if it is true, $(5, 10)$ is the only number/base pair with this property (excluding $n=1$). It might even be the only number/base pair with these properties minus the form restriction, but I haven't yet worked that out.)

Answer 1

Yes, this holds for all powers of 5.

Given a positive integer $n$, $5^{n}$ has $\lceil n\log_{10}5\rceil$ digits. The last one of these is a $5$, and we’ll need it to write the exponentiation. The question then becomes: is it possible to write $n$ using the remaining $\lfloor n\log_{10}5\rfloor$ digits?

Out of $k$ decimal digits, it’s always possible to form at least $(k - 10) / 2$ pairs of identical digits.

Ones can be formed from any pair of identical digits: $1 = 0^{0}$, and $1 = d/d$ for $d \neq 0$.

Any number can be written using only ones, by expanding its binary representation. For example: $$42 = 2^{5} + 2^{3} + 2 = (1+1)(1+1)(1+1)(1+1)(1+1) + (1+1)(1+1)(1+1) + (1+1)$$

Writing $n$ in this way requires at most $(\log_2n)^{2} + 1$ ones. We can then get rid of all the remaining digits, by using $(1-1)\times (\text{whatever})$.

For all $n\geq 165$, $(\lfloor n\log_{10}5\rfloor - 10) / 2 \geq (\log_2n)^{2} + 1 + 2$, making it possible to use this scheme. It's then easy to check $n < 165$ individually.

It might even be the only number/base pair with these properties minus the form restriction, but I haven't yet worked that out.

This is trivially false. $(25, 10)$ is another number/base pair with the same properties, minus the form restriction.