If I have to be completely rigorous, I believe I would have to prove that if $n=2^{e_1}3^{e_2}5^{e_3}...$ is $n$'s prime factorization, then $(n)_{10}$ ends in exactly $k = \min\{e_1,e_3\}$ zeros.
But this is a base-dependent property, right? I can't just assume it will hold in every base.
If I assume that $(n)_{10}$ ends in $k$ zeros, I'm assuming the following:
$$(n)_{10} = r_n10^n+r_{n-1}10^{n-1}+\dots+r_{n-k}10^{n-k} = \frac{2^n 5^n}{5^k} \bigg( r_n 5^k + r_{n-1}\frac{5^{k-1}}{2}+\dots+r_{n-k}\frac{5}{2^k} \bigg)$$
with $r_i \neq0$, for every $r_i$.
But I got stuck there. Do I have to make something similar to $(n)_5$ appear there or what? Is this solvable using the Binomial Theorem?
$(n)_p$ ends in $k$ zeroes iff $p^k || n$ where $p$ is a prime. In your case $(n)_ {10} $ ends with $k$ zeroes . So $10^k | n \implies 5^ k |n$ Thus $(n)_5$ ends in at least $k$ zeroes