Let $p$ be a prime and let $n$ be a positive integer. Then $p$ divides $\binom{2n}{n}$ exactly $\sum_{i=1}^{\lfloor \log_p 2n\rfloor} \left\lfloor\frac{2n}{p^i}\right\rfloor-2\left\lfloor\frac{n}{p^i}\right\rfloor$ times. This follows from $v_p(n!) = \sum_{i=1}^{\lfloor \log_p n \rfloor}\left\lfloor\frac{n}{p^i}\right\rfloor$, where $v_p(n)$ is the largest power of $p$ that exactly divides $n$, and $\binom{2n}{n}=\frac{(2n)!}{n!n!}$. So, why then does $p^r\vert\binom{2n}{n}$ imply that $p^r\leq2n$?.
Edit: Thank you to @BrunoB. I see why now. From the definition of the greatest integer function you have $$\left\lfloor\frac{2n}{p^i}\right\rfloor-\left\lfloor\frac{n}{p^i}\right\rfloor\leq\frac{2n-n}{p^i}=\frac{n}{p^i}\lt \left\lfloor\frac{n}{p^i}\right\rfloor+1$$ $$\left\lfloor\frac{2n}{p^i}\right\rfloor-\left\lfloor\frac{n}{p^i}\right\rfloor\lt\left\lfloor\frac{n}{p^i}\right\rfloor+1 $$ $$\left\lfloor\frac{2n}{p^i}\right\rfloor-2\left\lfloor\frac{n}{p^i}\right\rfloor\lt1 $$
Since every term in the summation is less than 1, and there are at most $\left\lfloor\log_p2n\right\rfloor$ terms, you have $$v_p\left(\binom{2n}{n}\right)\lt\left\lfloor\log_p2n\right\rfloor\leq\log_p2n$$ $$p^{v_p\left(\binom{2n}{n}\right)}\leq 2n$$
Since $p^r\vert\binom{2n}{n}$, $r$ cannot exceed $v_p\left(\binom{2n}{n}\right)$. Thus, $$p^r \leq p^{v_p\left(\binom{2n}{n}\right)}\leq 2n$$ $$p^r\leq 2n$$ $\square$