How to prove $\sum\limits_{k\mathop=0}^n\binom nk\cdot\sum\limits_{k\mathop=0}^n\binom nk=\sum\limits_{k\mathop=0}^n\binom{2n}k$ I need to figure out what could be a suitable combinatoric explanation
that fits both sides of the equation and why:

$$\left(\sum_{k\mathop=0}^n\binom nk\right)\cdot\left(\sum_{k\mathop=0}^n\binom nk\right)=\sum_{k\mathop=0}^n\binom{2n}k$$

 A: The total number of ways to choose objects from a pile of $2n$ objects
is the same as the total number of ways to choose objects from each of two piles of $n$ objects.
A: Hint: $((1+x)^n)^2=(1+x)^{2n}$ with $x=1$.
A: Something is wrong with your identity:
Since $\binom{2n}k=\binom{2n}{2n-k}$, we have 
$$S=\sum_{k\mathop=0}^n\binom{2n}k=\sum_{k\mathop=n}^{2n}\binom{2n}k.$$ Therefore, $$2S=\sum_{k\mathop=0}^{2n}\binom{2n}k 1^k 1^{2n-k}+\binom{2n}n=2^{2n}+\binom{2n}n,$$ $$S=2^{2n-1}+\frac{1}{2}\binom{2n}n$$ where we used binomial formula.
On the other hand $$\left(\sum_{k\mathop=0}^n\binom nk\right)\cdot\left(\sum_{k\mathop=0}^n\binom nk\right)=2^n 2^n=2^{2n}.$$
As you can see theese two things are not the same.
You probably wanted to show $$\left(\sum_{k\mathop=0}^n\binom nk\right)\cdot\left(\sum_{k\mathop=0}^n\binom nk\right)=\sum_{k\mathop=0}^{2n}\binom{2n}k.$$ This holds since the left side is equal to $2^n 2^n$ and the right side is equal to $2^{2n}$, which goes straightforward by using binomial formula.
