Derivation of binomial coefficient in binomial theorem.
Solution 1:
There are two possible questions here: (i) Why is $\binom{n}{k}$ ("$n$ choose $k$") given by the expression that you gave and (ii) Why is the coefficient of $x^k$ in the expansion of $(1+x)^n$ equal to $\binom{n}{k}$.
We answer question (i), giving two combinatorial arguments. For some reason, the first has been more standard in textbooks than the second.
First argument: Note that $\binom{n}{k}$ is the number of ways to choose $k$ people from $n$ people. We look at a different problem. How many ways are there to choose $k$ people and line them up in a row? We will calculate this number $N$ in two different ways.
Way $1$: Choose the $k$ people. This by definition can be done in $\binom{n}{k}$ ways. For every way of choosing the people, there are $k!$ ways to line them up. It follows that $$N=\binom{n}{k}k!.$$ Of course, we officially don't know a formula for $\binom{n}{k}$. But we soon will!
Way $2$: The first person for the lineup can be chosen in $n$ ways. For every one of these ways, the second person in the line can be chosen in $n-1$ ways. Continue, for a total of $k$ times. It follows that $$N=(n)(n-1)(n-2)\cdots(n-k+1)$$ Multiply "top" and "bottom" (which is invisible, it is $1$) by $(n-k)!$. We conclude that $$N=\frac{n!}{(n-k)!}$$
Way $1$ and Way $2$ of counting are both correct, so the answers must be the same. It follows that $$\binom{n}{k}k!=\frac{n!}{(n-k)!}.$$ Now dividing both sides by $k!$ gives us the formula you mentioned.
Second argument: We count in two different ways the number of ways to arrange $n$ people in a row. Let $n$, for example, be $100$.
First way: It is a standard fact that this number is $n!$, or in our concrete case, $100$!.
Second way: We count the same thing in a somewhat strange way. Pick once and for all an integer $k$, with $0\le k\le n$. In our concrete case with $n=100$, pick say $k=17$.
Choose the collection of people who will occupy the first $k$ positions. This can be done in $\binom{n}{k}$ ways. For each such choice, the $k$ chosen people can be lined up in $k!$ ways. And for every way of choosing and lining up the $k$ chosen people, there are $(n-k)!$ ways to line up the unchosen $n-k$ people in the $n-k$ positions in back of all the chosen people, for a total of $$\binom{n}{k}k!(n-k)!$$ ways. In the concrete case, it would be $\binom{100}{17}17!83!$.
Finally, compare the two answers. We conclude that $$n!=\binom{n}{k}k!(n-k)!,$$ and now dividing both sides by $k!(n-k)!$ gives the desired result.
Remark: Counting something in two different ways is a powerful method for obtaining combinatorial identities. There are many examples.
Solution 2:
I don't know how the result was originally derived, but here is one calculation.
Let $f(x) = (1+x)^n$ and let us try and find the Maclaurin series for $f(x)$.
The $0$-th derivative $f^{(0)}(x)$ is just $f(x) = (1 + x)^n$ itself. The first derivative is $f^{(1)}(x) = n(1+x)^{n-1}$, the second is $f^{(2)}(x) = n(n-1)(1+x)^{n-2}$, and so on. The $k$-th derivative is $f^{(k)}(x) = n(n-1)\cdots (n-k+1)(1+x)^{n-k}$, and finally the $n$-th derivative is $f^{(n)}(x) = n(n-1)\cdots 2 \cdot 1 = n!$ which is a constant. Hence, the first $n+1$ terms of the Maclaurin series for $f(x)$ are $$f(x) = (1 + x)^n \approx \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k = \sum_{k=0}^n \frac{n(n-1)\cdots (n-k+1)}{k!} x^k.$$ In fact, since $f^{(n)}(x)$ is a constant, $f^{(k)}(x) = 0$ for all $k > n$. Thus, the above is the complete Maclaurin series for $f(x)$; there is no approximation.
The Maclaurin series for $(1+x)^n$ is called the binomial theorem expansion of $(1+x)^n$, and the coefficient of $x^k$ in the series called a binomial coefficient. We denote this coefficient by $\binom{n}{k}$ and write $$f(x) = (1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$ Note that upon multiplying $\binom{n}{k}$ by $\frac{(n-k)!}{(n-k)!} = 1$ that $$\binom{n}{k}\times \frac{(n-k)!}{(n-k)!} = \frac{n(n-1)\cdots (n-k+1)}{k!}\times \frac{(n-k)!}{(n-k)!} = \frac{n!}{k!(n-k)!}.$$