Let's first count the number of bad paths, i.e. those paths which go beyond the line $y=x$.

Assuming $P$ is such a path. Given it goes beyond the line $y=x$, $P$ must touch some points on the line $y=x+1$. Let the first such intersection point be $M$. Then construct a mirroring path from $(0,0)$ to $M$ with respect to the line $y=x+1$ and keep the remaining unchanged. Name the new path as $P'$.

This is shown in the diagram below.

The mirroring point of the point $(0, 0)$ is $(-1, 1)$, so $P'$ is a path from $(-1, 1)$ to $(n, n)$. Every bad path $P$ will have a unique mirroring path $P'$. Conversely, every path $P'$ from $(-1, 1)$ to $(n, n)$ will have a unique mirroring path $P$ from $(0, 0)$ to $(n, n)$. Such a mirrored path $P$ must be bad because $P$ and $P'$ share a point $M$ on $y=x+1$ which means $P$ goes outside valid region.

Because the number of path $P'$ is $\binom{(n+1)+(n-1)}{n-1}$, therefore there this many bad paths $P$.

Meanwhile, there are totally $\binom{n+n}{n}$ paths from $(0,0)$ to $(n,n)$. It follows that the number of good paths equals $$\binom{2n}{n}-\binom{2n}{n-1} =\boxed{\frac{1}{n+1}\binom{2n}{n}}$$