Proving the Binomial Identity: A Shorter Solution Approach

[MathematicalPhysicist]

problem
prove that:
$$\forall n \in N \forall 0<=k<=2^{n-1} (C(2^n,k)=\sum_{j=0}^{k}C(2^{n-1},j)C(2^{n-1},k-j))$$

attempt at solution
induction seems to be too long I am opting for a shorter solution, so the sum that it's wrriten in the rhs is the square of the sum of the term C(2^(n-1),j) but other than that don't know how to procceed.

any advice?

thanks in advance.

 

[MathematicalPhysicist]

C is the binomial coefficient.

 

[MathematicalPhysicist]

so i guess no one here is that good with B.Cs?

 

[SanjeevGupta]

Have you tried using the fact that $$2^n=2^{n-1}+2^{n-1}$$
Then expand $$(1+x)^{2^n}$$ in two different ways, equating coefficients of $$x^k$$?