I like this question too.
\(25^\frac{1}{25}=5^{\frac{2}{5^2}}\\~\\ 125^\frac{1}{125}=5^{\frac{3}{5^3}}\\~\\ 625^\frac{1}{625}=5^{\frac{4}{5^4}}\\~\\ \)
\(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\\~\\ P = 5^{1/5} \cdot 5^{2/5^2} \cdot 5^{3/5^3} \cdot 5^{4/5^4} \dotsm\\~\\ P = 5^{{1/5}+2/5^2+{3/5^3} +{4/5^4} }\dotsm\\~\\ \)
So I need to find a sum to infinity.
\(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\dots +\\~\\ =\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dots +\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dots +\\~\\ =(\text{infinite sum of GP a=1/5, r=1/5})\dots \\ \qquad+(\text{infinite sum of GP a=1/25, r=1/5})\dots \\ \qquad+(\text{infinite sum of GP a=1/125, r=1/5})\dots +\\~\\ =\left[\frac{1}{5}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{25}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{125}\div(1-\frac{1}{5}))\right]+\dots \\ =\left[\frac{1}{5}\times\frac{5}{4}\right]+\left[\frac{1}{25}\times\frac{5}{4}\right]+\left[\frac{1}{125}\times\frac{5}{4}\right]+\dots \\ =\frac{1}{4}+\frac{1}{20}+\frac{1}{100}\dots\\~\\ \text{Infinite sum of a GP a=1/4 r=1/5}\\ =\frac{1}{4}\div(1-\frac{1}{5})\\~\\ =\frac{1}{4}\times \frac{5}{4}\\~\\ =\frac{5}{16} \)
\(P = 5^{1/5} \cdot 25^{1/25} \cdot 125^{1/125} \cdot 625^{1/625} \dotsm\\ P= 5^\frac{5}{16}\)
NOTE:
I wrote the latex and worked the logic at the same time so there is plenty of room for careless error.
You need to work through (understand) what I have done and find any errors that might be there.
Unless you only want the answer of course in which case you may or may not get lucky.
LaTex:
\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\dots +\\~\\
=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}\dots +\frac{1}{5^3}+\frac{1}{5^4}+\frac{1}{5^5}\dots +\\~\\
=(\text{infinite sum of GP a=1/5, r=1/5})\dots \\
\qquad+(\text{infinite sum of GP a=1/25, r=1/5})\dots \\
\qquad+(\text{infinite sum of GP a=1/125, r=1/5})\dots +\\~\\
=\left[\frac{1}{5}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{25}\div(1-\frac{1}{5}))\right]+\left[\frac{1}{125}\div(1-\frac{1}{5}))\right]+\dots \\
=\left[\frac{1}{5}\times\frac{5}{4}\right]+\left[\frac{1}{25}\times\frac{5}{4}\right]+\left[\frac{1}{125}\times\frac{5}{4}\right]+\dots \\
=\frac{1}{4}+\frac{1}{20}+\frac{1}{100}\dots\\~\\
\text{Infinite sum of a GP a=1/4 r=1/5}\\
=\frac{1}{4}\div(1-\frac{1}{5})\\~\\
=\frac{1}{4}\times \frac{5}{4}\\~\\
=\frac{5}{16}
=
!
