<s>[INST] Prove that if $p^{a}$ is a factor of the canonical factorization of ${{2n}\choose{n}}$ then $p^{a} < 2n$? Prove that if $p^{a}$ is a factor of the canonical factorization of ${{2n}\choose{n}}$ then $p^{a} < 2n$?
My attempt:
$${{2n}\choose{n}} = \frac{(2n)!}{n!n!}$$
Let $a_1$ be the highest of power of $(2n)!$
Let $a_2$ be the highest of power of $n!$
So the highest power of $\frac{(2n)!}{n!n!}$ = $a_1 - 2a_2$
where $a_1 <= 2n - 1$ and $2a_2 <= 2n - 2$
Therefore the highest power of $p$ that divides $\frac{(2n)!}{n!n!}$ is $2n - 1 - 2n + 2 = 1$.
Since $a <= 1 \implies p^{a} < 2n$
Am I in the right track? Any idea?  
Update 
Following Ross Millikan's hint:
Let $a$ be the highest power of $p$ such that $p^{a}|n!$
Then, $a$ =  $\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor +\lfloor \frac{n}{p^3} \rfloor + \ldots \lfloor \frac{n}{p^k} \rfloor$
Let $b$ be the highest power of $p$ such that $p^{b}|(2n)!$
Then, $b$ =  $\lfloor \frac{2n}{p} \rfloor + \lfloor \frac{2n}{p^2} \rfloor +\lfloor \frac{2n}{p^3} \rfloor + \ldots \lfloor \frac{2n}{p^q} \rfloor$
$\Longrightarrow b - 2a$ is the highest power of $p$ such that $p|\frac{(2n)!}{n!n!}$  
Where $a, b \in N \implies b - 2a < b$
Besides, $p^{b} < 2n$
$\therefore p^{b - 2a} < p^{b} < 2n$
Am I in the right track now?  
Thanks,
Chan
 [/INST] Hint: The highest power of a prime,$p$, that divides $n!$ is $\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor +\lfloor \frac{n}{p^3} \rfloor + \ldots \lfloor \frac{n}{p^k} \rfloor$.  This is your $a_2$.  Can you compare twice this with the expression for $2n$, your $a_1$?
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