How can I prove that the binomial coefficient ${n \choose k}$ is monotonically nondecreasing for $n \ge k$? I want to prove that the binomial coefficient ${n \choose k}$ for $n \ge k$ is a monotonically nondecreasing sequence for a fixed $k$. How do I do this?
 A: Fix $k \geq 0$.  Show that $${n+1 \choose k} = \frac{(n+1)!}{k!(n+1-k)!} \geq \frac{n!}{k!(n-k)!} = {n \choose k}.$$
But this follows whenever $n+1 \geq n+1 - k$, so...
A: From the recursive formula for binomial coefficient
$$
\binom{n}{k} = \binom {n-1}{k} + \binom{n-1}{k-1} \qquad (n, k > 0), 
$$
it is clear that $\binom{n}{k} \geqslant \binom{n-1}{k}$.

The claim is even more obvious when one thinks of the combinatorial interpretation of the binomial coefficient. Every $k$-subset of $\{ 1, 2, \ldots,  n-1 \}$ is also a $k$-subset of $\{1, 2, \ldots, n\}$; so it immediately follows that $\binom{n-1}{k} \leqslant \binom{n}{k}$.
A: If if you keep $k$ fixed and increase $n$ the value of $\binom{n}{k}$ will increase monotonically. 
$$
\begin{align*}
\binom{n}{k} &= \frac{n!}{k!(n-k)!}
\\ &= \frac{n(n-1)....(n-k+1)}{k!} \tag{1}
\end{align*}
$$
If you differentiate the RHS of the above relation wrt to $n$ keeping $k$ constant, you will get a positive [or zero derivative: zero derivative occurs for $k=0$] derivative for $n \geqslant k$ [both $n$ and $k$ positive]. Therefore $\binom{n}{k}$ is a monotonocally increasing function of $n$ when $k$ is kept constant.
You may consider the function expressed by $(1)$ as a continuous function of $n$ [$k$ is of course a fixed integer]. But our interest will be on the integral values of $n$. These values occupy discrete positions on the domain of continuous function considered.
For $k=0$, the function is defined to be a constant ($=1$). 
