Let's begin with:
[MATH](x+1)^n(1+x)^n=(x+1)^{2n}[/MATH]
Use the binomial theorem on each binomial...what do you get?
Using the binomial theorem, we get:
[MATH]\left(\sum_{k=0}^n\left({n \choose k}x^{n-k}\right)\right)\left(\sum_{k=0}^n\left({n \choose k}x^{k}\right)\right)=\sum_{k=0}^{2n}\left({2n \choose k}x^{2n-k}\right)[/MATH]
To make things a bit easier to see, let's rewrite the LHS as:
[MATH]\left({n \choose 0}x^{n}+{n \choose 1}x^{n-1}+\cdots+{n \choose n-1}x^{1}+{n \choose n}x^{0}\right)\left({n \choose 0}x^{0}+{n \choose 1}x^{1}+\cdots+{n \choose n-1}x^{n-1}+{n \choose n}x^{n}\right)[/MATH]
Now, we can easily see that on the LHS, we will have the term:
[MATH]\sum_{k=0}^n\left({n \choose k}^2\right)x^n[/MATH]
And on the RHS of the equation, we have the term:
[MATH]{2n \choose n}x^n[/MATH]
Hence, the following must be true:
[MATH]\sum_{k=0}^n\left({n \choose k}^2\right)={2n \choose n}[/MATH]