Can anybody give me a proof of binomial theorem that doesn't use mathematical induction? I have seen the proofs of Binomial theorem that use induction, but I would like to know if there is any other way to prove the theorem (apart from the combinatorial way that is already there).
 A: Any proof must have induction hidden inside. Anyhow, here is one "explicit" proof:
$$(x+y)^n=(x+y)(x+y)...(x+y)$$
Now, when we open the brackets, we get products of $x$ and $y's$. Every term the product of $k$ x' and $n-k$ y's. It follows that 
$$(x+y)^n=a_0x^n+a_1x^{n-1}y+...+a_kx^{n-k}y^k+...+a_ny^n$$
Now, what we need to figure is what is each $a_k$. $a_k$ counts how many times we get the term $x^{n-k}y^k$ when we open the brackets. 
We need to get $y$ from $k$ out of the $n$ brackets and this can be done in  $\binom{n}{k}$ ways. Now, the $x$ must come from the remaining brackets, we have no choices here. 
Thus $x^{n-k}y^k$ appears $\binom{n}{k}$ times, which shows 
$$a_k=\binom{n}{k}$$
this proves the formula.
A: $(a+b)^n = a^n + b^n + \text{ other terms}$, clearly.  Let's look at the $a^{n-1}b$ term.  Clearly there are $n$ ways to form that term given
$(a + b) \dots (a + b) \text { (n times )}$. 
So so far we have $(a + b)^n = a^n + b^n + na^{n-1} b + n b^{n-1}a + \dots $
Now consider $a^{n - k} b^{k}$ terms.  Those form from a $(a+b)^{k}$ and a $(a + b)^{n-k}$.  There are $\binom{n}{k}$ ways to choose such an $(a + b)^k$ out of the full expression $(a + b)^n$.  QED
Now, I used a lot of clearly's so let me know if I can make them more clear.
