TAOCP 1.2.6 Exercise 9
For every integer $n \ge 0$, Eq.
Section 1.2.6: Binomial Coefficients
Exercise 9. [01] What is the value of $\binom{n}{n}$? (Consider all integers $n$.)
Verified: yes
Solve time: 39s
For every integer $n \ge 0$, Eq. (5) gives
$$ \binom{n}{n} = \frac{n!}{n!(n-n)!} = \frac{n!}{n!,0!}. $$
Since $\binom{r}{0}=1$ by Eq. (4), Eq. (5) with $n=0$ implies $0!=1$; hence
$$ \binom{n}{n}=1. $$
If $n<0$, Eq. (3) defines $\binom{r}{k}=0$ whenever the lower argument $k$ is a negative integer. Taking $r=n$ and $k=n$, we obtain
$$ \binom{n}{n}=0 \qquad (n<0). $$
Therefore
$$ \boxed{\binom{n}{n}= \begin{cases} 1, & n\ge 0,\[4pt] 0, & n<0. \end{cases}} $$