How many zeroes are there at the end of $36!^{36!}$?

Question

Could you please tell me how many zeroes are there at the end of $36!$ to the power $36!$, i.e., $36!^{36!}$? I have been trying to find out. Read some reviews and answers related this but didn't understand at all. Thanks for the help.

Answer 1

$\sum\limits_{k:5^k<n}\lfloor\frac{n}{5^k}\rfloor$ gives the number of trailing zeros in n!

Above suggests that there are eight trailing zeros in 36!, so we can write $36!=\text{some number ending with non zero digit, (say k)}\times 10^8$

$\implies 36!^{36!}=(k\times 10^8)^{36!}$

We just need to be concerned about the ${10^8}^{36!}$ part since $k^{36!}$ will not contribute any more zeroes (if last digit of k$\ne0$ then last digit of $k^m\ne0\forall m\in\mathbb{N}$

And hence it has $8.36!$ trailing zeros.

Answer 2

De Polignac's formula tells us that $36!$ there are 34 powers of $2$ and 8 powers of $5$, so when $36!$ is factored, there are $8$ zeros because there is a $10^8$ in its factorisation. Since each power of $36!$ adds 8 zeroes, $36!^{36!}$ has $8\cdot36!$ zeros at the end.