For which value of n ∈ N ,   n ! has 13 trailing zeros?

We know in $n!$ exponent of $5$ is always less than or equal to exponent of $2$

Hence, exponent of $5$ is equal to the number of trailing zeroes

We know, exponent of $5$ in $n!$ is given as 

$\Rightarrow 13=\left[\frac{n}{5}\right]+\left[\frac{n}{5^2}\right]+\left[\frac{n}{5^3}\right]+...$

Using $n=57$ from the options, we get 

$=\left[\frac{57}{5}\right]+\left[\frac{57}{25}\right]+\left[\frac{57}{125}\right]$

$=11+2+0 ...$

So, in $57!$ the power of $5$ is $13$