The number of zeros at the end of $70!$ is

$70!=2^a\times 3^b\times 5^c\times 7^d\times \mathrm{...}$

Now,

     $a=E_2(70!)=\left[\frac{10}{2}\right]+\left[\frac{70}{2^2}\right]+\left[\frac{70}{2^4}\right]+\left[\frac{70}{2^5}\right]+\left[\frac{70}{2^6}\right]$

$\Rightarrow$   $a=35+17+8+4+2+1=67$

And,  

$C=E_5(70!)=\left[\frac{70}{5}\right]+\left[\frac{70}{5^2}\right]=14+2=16$

$\therefore$   $70!=2^{67}\times 5^{16}\times 3^b\times 7^d\times \mathrm{...}$

$\Rightarrow$    $70!={(2\times 5)}^{16}\times 2^{51}\times 3^b\times 7^d\times \mathrm{...}$

$\Rightarrow$    $70!={10}^{16}\times 2^{51}\times 3^b\times 7^4\times \mathrm{...}$

Thus, the number of zeros at the end of $70!$ is 16.