How to find the highest power of a prime $p$ that divides $\prod \limits_{i=0}^{n} 2i+1$? 
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? 

Given an odd prime $p$, how does one find the highest power of $p$ that divides
$$\displaystyle\prod_{i=0}^n(2i+1)?$$
I wrote it down all paper and realized that the highest power of $p$ that divides this product will be the same as the highest power of $p$ that divides $(\lceil\frac{n}{2}\rceil - 1)!$
Since 
$$10! = 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10$$ while 
$$\prod_{i=0}^{4} (2i+1) = 1\times 3\times 5\times 7\times 9$$ 
Am I in the right track?  
Thanks,
Chan
 A: Note that $\displaystyle \prod_{i=1}^{n} (2i-1) = \frac{(2n)!}{2^n n!}$.
Clearly, the highest power of $2$ dividing the above product is $0$.
For odd primes $p$, we proceed as follows.
Note that the highest power of $p$ dividing $\frac{a}{b}$ is nothing but the highest power of $p$ dividing $a$ - highest power of $p$ dividing $b$.
i.e. if $s_p$ is the highest power of $p$ dividing $\frac{a}{b}$ and $s_{p_a}$ is the highest power of $p$ dividing $a$ and $s_{p_b}$ is the highest power of $p$ dividing $b$, then $s_p = s_{p_a}-s_{p_b}$.
So the highest power of $p$ dividing $\displaystyle \frac{(2n)!}{2^n n!}$ is nothing but $s_{(2n)!}-s_{2^n}-s_{n!}$.
Note that $s_{2^n} = 0$.
Now if you want to find the maximum power of a prime $q$ dividing $N!$, it is given by
$$s_{N!} = \left \lfloor \frac{N}{q} \right \rfloor + \left \lfloor \frac{N}{q^2} \right \rfloor + \left \lfloor \frac{N}{q^3} \right \rfloor + \cdots$$
(Look up this stackexchange thread for the justification of the above claim)
Hence, the highest power of a odd prime $p$ dividing the product is $$\left ( \left \lfloor \frac{2N}{p} \right \rfloor + \left \lfloor \frac{2N}{p^2} \right \rfloor + \left \lfloor \frac{2N}{p^3} \right \rfloor + \cdots \right ) - \left (\left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots \right)$$
