Binomial fun

It is known that $C_{i} = \binom{n}{i}$ . (1)

(This is because if one one writes out the $n$ separate $(1+x)$ s, then the $i$ th term can be created by picking exactly $i$ $x$ 's, of which there are $\binom{n}{i}$ ways of such choice.)

It is noteworthy that $C_{i} = \binom{n}{i} = \binom{n}{n-i} = C_{n-i}$ . (2)

$(1+x)^{2n} =( \sum_{i=0}^{n}C_{i})( \sum_{i=0}^{n}C_{i})$

Therefore $x^n$ is also formed by

$\sum_{i=0}^{n} C_{i} x^i \cdot C_{n-i} x^{n-i}$

$=\sum_{i=0}^{n} C_{i}C_{n-i} x^n$

$=\sum_{i=0}^{n} C_{i}^2 x^n$ by (2)

By (1), the $n$ th term (i.e. the coefficient of $x^n$ ) of $(1+x)^{2n}$ is $\binom{2n}{n} = \frac{2n!}{(n!)^2}$

QED.