Prove that for $n\in\mathbb{N}$, $\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$ How to show that the following relation? : for $n\in\mathbb{N}$, $$\sum_{k=1}^{\infty}\left[\frac{n}{5^{k}}\right]=15\iff\left[\frac{n}{5}\right]=13.$$ It's not obvious to me. Can anyone help me? Thank you!

Observe that $\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^k}\right\rfloor=\nu_5(n!)$, where $\nu_5(n!)$ is the exponent of the largest power of $5$ that divides $n!$ (if you are not familiar with this, then see Legendre's formula).
Since this sum is $15$, thus  the highest power of $5$ that divides $n!$ is $15$. 
Consider $n!=(1\cdot 2 \cdot 3 \cdot 4 \cdot \color{red}{5})(6\cdot 7 \cdot 8 \cdot 9 \cdot \color{red}{10})\dotsb ((n-4) \cdot (n-3) \cdot (n-2) \cdot (n-1) \cdot \color{red}{n}).$
Each group of $((a+1)\cdot (a+2) \cdot (a+3) \cdot (a+4) \cdot (a+5))$ contributes a single power of $5$, until we reach $25$, where we get a contribution of $2$ towards the exponent of $5$. Likewise until we reach $50$ each such group of five numbers contribute only a single power of $5$ and so on. Thus the number $n$ must be such that
$$n!=\underbrace{(1\cdot 2 \cdot 3 \cdot 4 \cdot \color{red}{5})}_{5^1}\underbrace{(6\cdot 7 \cdot 8 \cdot 9 \cdot \color{red}{10})}_{5^1}\dotsb \underbrace{(21\cdot 22 \cdot 23 \cdot 24 \cdot \color{red}{25})}_{\color{green}{\boxed{5^2}}}\dotsb \underbrace{(61\cdot 62 \cdot 63 \cdot 64 \cdot \color{red}{65})}_{5^1} \dotsb ()$$
When $n=65$, that is the first time we will have $\nu_5(n!)=15$. Thus $65 \leq n < 70$ which implies $\left\lfloor\frac{n}{5}\right\rfloor=13.$
