What is the proof of the Binomial Theorem, other than the induction method? How can we find the expansion of binomails with indices like 2n, 3n, 4n..? 
*

*The proof of the binomial theorem is almost always given with induction. Is there any other way to proof the binomial theorem. I have been trying to find it, but have not yet done it.

*What is the expansion of $((1+x)^n)^2$. Is there any way of finding the expansions for similar formulas like $((1+x)^n)^3$, $((1+x)^4)^n$,........., $((1+x)^n)^n).$
Will simplifying the terms like $(1+x)^n*(1+x)^n$ and then multiplying the general terms of both the expansions help?
 A: For your first question we can also show it using the Taylor series formula
$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k\ .$$
Fix $n\in\mathbb{N}$ and let $f(x) = (1+x)^n$. Then $f$ is analytic (it is just a polynomial) and so we can apply the above formula. We only need to compute the $k$th derivative at $0$. For $k\leq n$
$$f^{(k)}(x)= n\times(n-1)\times(n-2)\times\cdots\times(n-k+1)\times (1+x)^{n-k} =\frac{n!}{(n-k)!}(1+x)^{n-k}\ ,$$
while for $k> n$ we have
$$f^{(k)}(x)=0\ .$$
Maybe you can say this step needs induction, but it's certainly more clear than the whole formula. Plugging in $x=0$ we see
$$f^{(k)}(0)=\begin{cases}\frac{n!}{(n-k)!}& k\leq n\\ 0 & k > n\end{cases}$$
Inserting this back into the Taylor series formula gives
$$f(x) = \sum_{k=0}^n \frac{n!}{(n-k)!k!}x^k = \sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}x^k$$
Edit: To answer your second question $((1+x)^n)^m = (1+x)^{nm}$ and so you can just replace all the $n$'s by $nm$'s in the binomial theorem to get the answer.
A: Here is  generating function approach. We show the validity of
\begin{align*}
(1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k\tag{1}
\end{align*}
by showing that left-hand side and right-hand side have the same exponential generating function.
\begin{align*}
\color{blue}{e^{(1+x)z}}
\end{align*}

We start with the exponential generating function of the left-hand side. We obtain
\begin{align*}
\sum_{n=0}^\infty \color{blue}{(1+x)^n}\frac{z^n}{n!}&=\sum_{n=0}^\infty\frac{\left((1+x)z\right)^n}{n!}\\
&=e^{(1+x)z}\tag{2.1}\\
&=e^{xz}e^z\tag{2.2}\\
&=\left(\sum_{k=0}^\infty \frac{(xz)^k}{k!}\right)\left(\sum_{l=0}^\infty\frac{z^l}{l!}\right)\tag{2.3}\\
&=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{x^k}{k!}\,\frac{1}{l!}\right)z^n\tag{2.4}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{x^k}{k!}\,\frac{1}{(n-k)!}\right)z^n\tag{2.5}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{n!}{k!}\,\frac{1}{(n-k)!}x^k\right)\frac{z^n}{n!}\tag{2.6}\\
&\,\,=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n\binom{n}{k}x^k}\right)\frac{z^n}{n!}\tag{2.7}\\
\end{align*}
and the claim (1) follows.

Comment:

*

*In (2.1) we use the definition of the exponential series.


*In (2.2) we apply the rule $e^{p+q}=e^pe^q$.


*In (2.3) we use again the definition as in (2.1).


*In (2.4) we rearrange according increasing powers of $z$.


*In (2.5) we eliminate $l$ by substituting $l\to n-k$.


*In (2.6) we expand numerator and denominator with $n!$.


*In (2.7) we use $\binom{n}{k}=\frac{n!}{k!(n-k)!}$.
