How many zeros are there at the end of the $A$ How many  zeros are there at the end of $A$
where
$$A=\binom{2013}{0}\times\binom{2013}{1}\times\binom{2013}{2}\times\cdots\cdots\times\binom{2013}{2013}$$
I know this numbers $2013!$  at the end of zeros
$$n=\left[\dfrac{2013}{5}\right]+\left[\dfrac{2013}{5^2}\right]+\left[\dfrac{2013}{5^3}\right]+\left[\dfrac{2013}{5^4}\right]$$
but for $A$
$$\binom{2013}{k}=\dfrac{2013!}{k!(2103-k)!}$$
then
$$A=\dfrac{(2013!)^{2014}}{(1!2!3!\cdots2013!)^2}$$ I can't,Thank you
 A: If $v(n)$ denotes the exponent of $2$ in the decomposition in prime numbers of $n$ and $w(n)$ the exponent of $5$, then then number of zeros at the end of $n$ in base $10$ is $\min(v(n),w(n))$.
For factorials, it is quite obvious that $v(n!)>w(n!)$ so it is enough to check the powers of $5$. However, here we have a quotient of factorials, so the powers of $2$ might (and actually will) play some role.
Let's check the power of $5$ first. The denominator of the last expression may be rewritten as
$$1!2!\cdots2013!=1^{2013}2^{2012}\cdots 2013^1$$
so 
$$w(A)=2014w(2013!)-2\sum_{k=1}^{2013}(2014-k)w(k)=\sum_{k=1}^{2013}(2k-2014)w(k),$$
the last equality being true by decomposing the factorial.
It is now clear that it is enough to compute $\sum_{k=1}^{2013}k w(k)$ and $\sum_{k=1}^{2013}w(k)$. The latter is just $w(2013!)=501$ (using the formula given in the question).
For the former, we can look at the beginning of the sum : $5+10+15+20+2\cdot25+30+\cdots$ which can be rewritten as sum of multiple of $5$ plus sum of multiple of $25$ and so on.
This gives the formula
$$\sum_{k=1}^{2013}k w(k)=5\sum_{k=1}^{402}k+25 \sum_{k=1}^{80}k+125 \sum_{k=1}^{16}k+625 \sum_{k=1}^{3}k = \sum_{i=1}^{\infty}5^i\sum_{k=1}^{[2013/5^i]}k$$
(the last being the most general form).
This is easily computed using the classical formula $\sum k=\frac{n(n+1)}2$, and we find $w(A)=4516$.
Doing the same for powers of $2$, we get that $v(A)=\sum_{k=1}^{2013}(2k-2014)v(k)=3820$ (the same formulas apply with $5$ replaced by $2$), so we see that the number of zeros is $3820$.
