Proof(incomplete):
There are $\frac {n!}{(n - k)!} k$- permutations. Each $k$-element subset can be listed in $k!$ ways. The number of ways to first choose a $k$ element subset and to list the elements of that subset is $\binom{n}{k} \cdot \frac {n!}{(n - k)!}$.
We are trying to prove what $\binom{n}{k} $ is, yet we are allowed to use it in the middle of its own proof? Please, explain what's up with that.
edit: $\binom{n}{k} \cdot \frac {n!}{(n - k)!}$ should be $\binom{n}{k} \cdot k!$.
It depends on how you have defined $\binom nk$. If the definition of $\binom nk$ is "the number of $k$-element subsets of an $n$-element set" then to evaluate "the number of ways to first choose a $k$ element subset and to list the elements of that subset," you have, by definition, $\binom nk$ ways to perform the first step (choose a $k$-element subset), and for each of those ways you have $k!$ choices for the second step (order in which you list the elements of the subset), ergo the total number of ways to perform both steps is $\binom nk \cdot k!$.
So we may not yet know how to write a formula for $\binom nk$, but we know what it means and can use it in a larger formula.
Of course if we defined $\binom nk = \frac{n!}{(n-k)!k!}$ instead, then we wouldn't have to prove that fact, but we could use a similar argument to show that $\binom nk$ is the number of $k$-element subsets of an $n$-element set.
But if we defined $\binom nk$ as a binomial coefficient then we have some other work to do to prove that it is equal to either of the other two things.
To derive the formulas for$~_n C_k=\binom{n}{k}$ using the definition that $\binom{n}{k}$ represents the number of different $k$ element subsets out of a set of $n$ distinct elements where order does not matter, you may consider this a multiplication principle question.
Step 1: pick the first element to appear ($n$ choices), Step 2: pick the second element to appear ($n-1$ choices), ... Step $k$: pick the $k$th element to appear ($n-k+1$ choices).
Noticing then that we overcounted and for each "different" possibility, we have counted the possibility a total of $k!$ number of times (since those $k$ steps could have been done in any different order), we divide by the amount $k!$ to account for the symmetry involved.
Thus, $\binom{n}{k} = \dfrac{n\cdot(n-1)\cdot(n-2)\cdots(n-k+1)}{k!} = \dfrac{n!}{(n-k)!k!}$
As for the condition that $0\leq k\leq n$, note that there is no way to choose more things from our set than there exist elements in the set in the first place. Similarly, it doesn't make sense to take a negative amount of things from the set. Indeed, $m!$ is not defined for any negative integer $m$, and as such if $k$ is outside of the range $[0,n]$ then one of $k!$ or $(n-k)!$ would not be defined. In that case, we define $\binom{n}{k}$ to be $0$.
