Proof that binomial coefficient is a natural number 
Possible Duplicate:
Proof that a Combination is an integer 

What is the proof that the binomial coefficient is a natural number?
$$k\ge0,n\ge k \implies {n \choose k} \in N,$$ 
I guess it's a pretty simple one, but I just can't find it.
 A: Induction Hint:
$$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$$
$$\qquad \binom{a}{a}=1$$
A: The simplest argument is to show that $\binom{n}k$ is the number of $k$-element subsets of an $n$-element set; since this is obviously a non-negative integer the result is immediate.
To see this, note first that there are clearly $$n(n-1)(n-2)\dots(n-k+1)\tag{1}$$ ways to choose a specific sequence of $k$ distinct elements of an $n$-element set: the first may be any of the $n$ elements, the second may be any of the $n-1$ remaining elements, and so on. It’s convenient to rewrite the expression in $(1)$ as $$\frac{n!}{(n-k)!}\;.\tag{2}$$
Now observe that each $k$-element subset of the original $n$-element set can be arranged in $k!$ different sequences; the argument is exactly the same as the one used to justify $(1)$ above. Thus, expressions $(1)$ and $(2)$ actually count each $k$-element subset $k!$ times, once for each of its permutations. Consequently, the actual number of $k$-element subsets is $$\frac1{k!}\cdot\frac{n!}{(n-k)!}=\frac{n!}{k!(n-k)!}=\binom{n}k\;.$$
A: This is obvious from the binomial theorem: $(1+X)^n = \sum_{k=0}^n \binom{n}{k} X^k$.
