Find a simple formula for $\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+...+\binom{n}{n-1}\binom{n}{n}$ 
$$\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+...+\binom{n}{n-1}\binom{n}{n}$$

All I could think of so far is to turn this expression into a sum. But that does not necessarily simplify the expression. Please, I need your help.

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Note that
$\ds{{n \choose k} + {n \choose k + 1} = {n + 1 \choose k + 1}}$.

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{n - 1}{n \choose k}{n \choose k + 1}} =
\sum_{k = 0}^{n - 1}
{\bracks{{n \choose k} + {n \choose k + 1}}^{\,2} -
{n \choose k}^{2} - {n \choose k + 1}^{2}\over 2}
\\[5mm] = &\
{1 \over 2}\sum_{k = 0}^{n - 1}{n + 1 \choose k + 1}^{2} -
{1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k}^{2} -
{1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k + 1}^{2}
\\[5mm] = &\
{1 \over 2}\sum_{k = 1}^{n}{n + 1 \choose k}^{2} -
{1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k}^{2} -
{1 \over 2}\sum_{k = 1}^{n}{n \choose k}^{2}
\\[5mm] = &\
{1 \over 2}\bracks{\sum_{k = 0}^{n + 1}{n + 1 \choose k}^{2} -2} -
{1 \over 2}\bracks{\sum_{k = 0}^{n}{n \choose k}^{2} - 1} -
{1 \over 2}\bracks{\sum_{k = 0}^{n}{n \choose k}^{2} - 1}
\\[5mm] = &\
{1 \over 2}{2n + 2 \choose n + 1} - {2n \choose n}
\end{align}
where I used the well known result
$\bbx{\ds{\quad\sum_{i = 0}^{m}{m \choose i}^{2} = {2m \choose m}}}$.

\begin{align}
& \mbox{Moreover,}\quad
\bbox[10px,#ffd]{\sum_{k = 0}^{n - 1}{n \choose k}{n \choose k + 1}} =
{1 \over 2}\,{\pars{2n + 2}\pars{2n + 1}\pars{2n}! \over
\pars{n + 1}n!\pars{n + 1}n!} - {2n \choose n}
\\[5mm] = &\
\pars{{2n + 1 \over n + 1} - 1}{2n \choose n} =
{n \over n + 1}{2n \choose n} =
{n \over n + 1}{\pars{2n}! \over n!\,n!}
\\[5mm] = &\
{\pars{2n}! \over \pars{n + 1}!\pars{n - 1}!} =
\bbx{\ds{2n \choose n + 1}}
\\ &\ \mbox{}
\end{align}
