How do we handle the factorials in the binomials/choice numbers?

Question

Apparently the following is a known equality:

$\frac{1}{n + 1} {2n \choose n} = \frac{2n!}{n!(n + 1)!} = \frac{1}{2n + 1}{2n + 1 \choose n}$

but I can't really figure out how to produce the equality.
E.g.

$\frac{1}{n + 1} {2n \choose n} = \frac{1}{n + 1} \frac{(2n)!}{n! (2n - n)!} = \frac{1}{n + 1} \frac{2n (2n - 1)!}{n!n!} = \frac{1}{n + 1}\frac{2n (2n - 1)!}{n(n-1)!n!}= \frac{1}{n + 1}\frac{2 (2n - 1)!}{(n-1)!n!}$
but then I am not sure how to proceed. Similar for the reverse direction.

How are the factorials here properly handled and we get this equality?

Answer 1

From left to right:

$\frac{1}{n+1}\binom{2n}{n} = \frac{1}{n+1}\frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{(n+1)!n!} = \frac{1}{2n+1}\frac{(2n+1)!}{(n+1)!n!} = \frac{1}{2n+1}\binom{2n+1}{n} $