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..?

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.