How many zeros in the decimal representation of $5^n$? I'm curious about some properties of the powers of 5
$$5^2=25,\quad5^3=125,\quad 5^4=625,\quad 5^5=3125,\quad ...$$
Is it true that at least $50$% of the digits in the decimal representation of $5^n$ are non-zero? This seems pretty modest since assuming each digit will be equally likely, only about $10$% of the digits will be zero on average. The first zero occurs at $5^8=390625$ and the power with the largest percentage of zeros seems to be
$$5^{45}=28421709430404007434844970703125$$
in which $\approx 22$% of the digits are zero. I checked up to $5^{1000}$.
The difficulty is the statement seems so obvious from a probabilistic perspective, yet I can't pin down any definite theorems! 
Obviously the first and the last 3 digits will always be non-zero so we have at least 4 non-zero digits. But I'm hoping to prove some properties about $5^n$ in general which require the non-zero digits to be at least linear in $n$. So $50$% would be more than enough. Really any probability $\epsilon>0$ will do -- bigger the better though.
Maybe analyze $\langle 5 \rangle^\times$ in $\mathbb{Z}/10^k\mathbb{Z}$? Could probability theory produce the bound in question?
 A: A good picture is worth a thousand words. It seems that the number of zeroes is growing in an almost linear fashion with a fair amount of "noise" around the meadian line. Here are the results for a number of zeroes in $5^n$ up to $n=10,000$

Not much will change if you extend the range to $n=20,000$:

Linear fit gives the following approximation:
$$n_{zero}=0.0699383 n-0.606536$$
...which means that the number of zeroes is roughly around 7%.
Mathematica provides the following regression analysis:
$$\begin{array}{l|llll}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{t-Statistic} & \text{P-Value} \\
\hline
 1 & -0.606536 & 0.354669 & -1.71015 & 0.0872544 \\
 x & 0.0699383 & 0.0000307141 & 2277.07 & 5.874\cdot 10^{-24155} \\
\end{array}$$
If you try to fit the data with a parabola you get the following approximation:
$$n_{zero}=0.0931693 + 0.0697284 n+1.0494\cdot 10^{-8}n^2$$
In the given range, the quadratic term is almost neglectable which supports the conjecture that the relation between the exponent and the number of zeroes is approximately linear.
EDIT: Fun fact: $5^{58}$ has not a single zero.
EDIT 2: Mathematica code to play with:
countZeros[n_] := Module[
   {m, cnt, d},
   m = n;
   cnt = 0;
   While[m > 0,
    d = Mod[m, 10];
    If[d == 0, cnt++];
    m = Quotient[m, 10];
    ];
   Return[cnt];
   ];

analyzeExponents[limit_] := Module[
   {i, data, exp},
   i = 0;
   exp = 1;
   data = {};
   While[i <= limit,
    i++;
    exp *= 5;
    AppendTo[data, {i, countZeros[exp]}];
    If[Mod[i, 100] == 0, Print["Reached i=", i]];
    ];
   Return[data];
   ];

ListPlot[analyzeExponents[10000]]

