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