[proofplan]
We split according to whether the integer $k$ lies in the admissible range $0 \leq k \leq n$. If $k$ is outside this range, then $n-k$ is outside the same range on the opposite side, so both binomial coefficients are zero by convention. If $0 \leq k \leq n$, the factorial formula gives identical expressions after commuting the two denominator factors.
[/proofplan]
[step:Show that the zero convention gives equality outside the admissible range]
Assume first that $k < 0$. Then $n-k > n$, since $-k > 0$. By the stated zero convention,
\begin{align*}
\binom{n}{k} = 0
\end{align*}
and
\begin{align*}
\binom{n}{n-k} = 0.
\end{align*}
Hence $\binom{n}{k} = \binom{n}{n-k}$.
Assume next that $k > n$. Then $n-k < 0$. Again by the stated zero convention,
\begin{align*}
\binom{n}{k} = 0
\end{align*}
and
\begin{align*}
\binom{n}{n-k} = 0.
\end{align*}
Hence $\binom{n}{k} = \binom{n}{n-k}$ in this case as well.
[guided]
The only issue outside the usual factorial range is that the expression $\binom{n}{k}$ is not being evaluated by factorials. Instead, the theorem explicitly gives a convention: any lower index outside the interval from $0$ to $n$ gives value $0$.
First suppose $k < 0$. Then $k$ is outside the admissible range, so
\begin{align*}
\binom{n}{k} = 0.
\end{align*}
The reflected lower index is $n-k$. Since $k < 0$, we have $-k > 0$, and therefore $n-k = n+(-k) > n$. Thus $n-k$ is also outside the admissible range, now above $n$, so
\begin{align*}
\binom{n}{n-k} = 0.
\end{align*}
Both sides are zero, hence they are equal.
Now suppose $k > n$. Then $k$ is outside the admissible range above $n$, so
\begin{align*}
\binom{n}{k} = 0.
\end{align*}
The reflected lower index satisfies $n-k < 0$, so it is outside the admissible range below $0$. Therefore
\begin{align*}
\binom{n}{n-k} = 0.
\end{align*}
Again both sides are zero, so
\begin{align*}
\binom{n}{k} = \binom{n}{n-k}.
\end{align*}
[/guided]
[/step]
[step:Use the factorial formula in the admissible range]
It remains to consider the case $0 \leq k \leq n$. Then $n-k$ is an integer satisfying $0 \leq n-k \leq n$, so both lower indices are admissible. By the factorial formula for binomial coefficients,
\begin{align*}
\binom{n}{k} = \frac{n!}{k!(n-k)!}.
\end{align*}
Using commutativity of multiplication in the denominator,
\begin{align*}
\frac{n!}{k!(n-k)!} = \frac{n!}{(n-k)!k!}.
\end{align*}
Applying the factorial formula again to the admissible index $n-k$ gives
\begin{align*}
\frac{n!}{(n-k)!k!} = \binom{n}{n-k}.
\end{align*}
Therefore $\binom{n}{k} = \binom{n}{n-k}$.
[/step]
[step:Combine the cases]
The three alternatives $k < 0$, $0 \leq k \leq n$, and $k > n$ exhaust all integers $k \in \mathbb{Z}$. In each case we have proved
\begin{align*}
\binom{n}{k} = \binom{n}{n-k}.
\end{align*}
This proves the claimed symmetry of binomial coefficients.
[/step]
