Show that the distribution of zeroes across all digits of all n! for $n\in \mathbb{N}$ converges to $ \frac{1}{6} $ and hence, $\frac{5}{54}$ for all other digits 1 through 9.
In other words, let's say we take the factorial of 1,2,3 and 4. We get the digits 1,2,2,4 and 6. In this case the probability of 2 would be $\frac{2}{5}$. We can keep doing this all the way to infinity.
I wrote a python script and checked from 0 to 10000, and all probabilities converge to 0.09259259259.. = $\frac{5}{54}$ except zero, which converges to 0.16666... = $\frac{1}{6}$.
Can you prove mathematically why this is the case?
I doubt this is true
Using Stirling's approximation, the number of decimal digits in $n!$ is about $\frac1{\log_e(10)}\left(n \log_{e}(n) -n +\tfrac12 \log_{e}\left(n \right)+ \tfrac12 \log_e(2\pi)\right)$
while the number of trailing $0$s on the right hand side of $n!$ is close to $\frac n4-1$
suggesting that the proportion of digits of $n!$ which are $0$ might be estimated as being of the order of $$ \frac1{10} + \frac{9}{10} \frac{\frac n4-1}{\frac1{\log_e(10)}\left(n \log_{e}(n) -n +\tfrac12 \log_{e}\left(n \right)+ \tfrac12 \log_e(2\pi)\right)}$$ with some natural variation about this
With $n=10000$ this estimate is about $0.163$, which may help explain your empirical result of $\frac16$
But as $n$ increases further, this estimate tends to $\frac1{10}$, essentially because $n\log_e(n)$ grows faster than $\frac n4$. For example with $n=10^{10}$ the estimate would be about $0.123$, while for $n=10^{100}$ it would be about $0.102$