Prove that $s_p(n-1)+s_p(n)+1-s_p(2n)\ge v_p(n)\cdot (p-1)$ for $p|n$ 
Prove that $\frac{(2n)!}{n!(n+1)!}$ is an integer.

Now, one way of proving this is $$\frac{(2n+1)!}{n!(n+1)!}=\frac{2n+1}{n}\cdot \frac{(2n)!}{(n-1)!(n+1)}=\frac{2n+1}{n}\cdot \binom{2n-1}{n+1}\in \Bbb Z$$
but $\gcd(n,2n+1)=1\implies n|\binom{2n}{n-1}$

The other possible way I think is using $v_p.$
Let $p$ be an odd prime dividing $n$ then $$v_p\left(\frac{(2n)!}{(n-1)!(n+1)!}\right)=v_p(2n!)-v_p(n-1!)-v_p(n+1!)=\frac{2n-s_p(2n)-n+1+s_p(n-1)-n-1+s_p(n+1!)}{p-1}=\frac{s_p(n-1)+s_p(n+1)-s_p(2n)}{p-1}$$
where $s_p$ denotes the sum of digits of $n$ is base $p$.
Since $p|n\implies s_p(n+1)=s_p(n)+1.$
And it is well known that $p-1|x-s_p(x).$
Now, to show that $n|\left(\frac{(2n)!}{n!(n+1)!}\right)$ enough to show that $$s_p(n-1)+s_p(n)+1-s_p(2n)\ge v_p(n)\cdot (p-1)$$
When $2|n$ note that $v_2(\frac{(2n)!}{n!(n+1)!})=v_2(\frac{(2n+1)!}{n!(n+1)!})=v_2(\binom{2n+1}{n})$
Any idea on how to prove this without using the first proof? Also, this identity is definitely true because of the first proof.
 A: We have the following basic inequality : for all $m,n \geq 0$ , we have $$s_p(m) + s_p(n) \geq s_p(m+n)$$ (with equality if and only if there are no "carries"). This is not particularly difficult to prove in isolation.

Using this, even if $p$ doesn't divide $n$, we can see that the inequality holds. In this case, $v_p(n) = 0$, so all we need to prove is that $s_p(n-1)+s_p(n)+1 \geq s_p(2n)$, but this is clear once you notice that $$
s_p(2n) \leq s_p(n+1)+s_p(n-1) \leq s_p\left(n\right) +s_p(1)+ s_p(n-1)
$$
and $s_p(1) = 1$.

We induct on $v_p(n)$ now. Suppose the fact is true for $v_p(n) = k$ with $k \geq 0$. We will assume that $v_p(n) = k+1$ so that $p|n$, and note that \begin{align}
&s_p(n-1)+ s_p(n)+1-s_p(2n) \\ &= \left[s_p\left(\frac{n}{p}-1\right) + (p-1)\right]+s_p\left(\frac{n}{p}\right)+1 -s_p\left(\frac{2n}{p}\right)\\
&= (p-1) + \left[s_p\left(\frac{n}{p}-1\right) + s_p\left(\frac np\right) + 1 - s_p\left(\frac{2n}{p}\right)\right]\\
&  \geq (p-1) + k(p-1) = (k+1)(p-1)
\end{align}
as desired.
