One of the most important and basic identities in binomial summations in Vandermonde's identity which states :
$$\sum_{k=0}^{r}\binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}$$
The identity can be generalized to the case when more than two binomial coefficients are multiplied.
Setting $m,r \mapsto n$ follows:
$$\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}=\binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}^2$$
Which is very well-known.
One of the questions related to this summation is :
$$\sum_{k=0}^{n}\binom{n}{k}^{4}=\sum_{k=0}^{n}\binom{\color{red}{n}}{\color{blue}{k}}\binom{\color{red}{n}}{\color{blue}{k}}\binom{\color{red}{n}}{\color{blue}{n-k}}\binom{\color{red}{n}}{\color{blue}{n-k}}$$
I thought that the answer would be $\binom{4n}{2n}$, but this is not right in general and I don't know why, since the sum of the red and blue parts is independent of the index (is a fixed number),so that's why I used the convolution (It would be highly appreciated if someone explain where was I wrong).
My question is : Does there exist any closed formula for :
$$\sum_{k=0}^{n}\binom{n}{k}^{r}$$
Where $r \in \mathbb N$.
One of the identities related to this question is Dixon’s Identity which has been mentioned here (page 3).
wolframalpha does not give a closed form.