Prove the number of words $w$ of length $n$ on the alphabet $[n]$ such that $w_{n−i} > i$ and $w_{i} ≤ w_{i+1} + 1$ is ${\frac 1{n+1}}{2n\choose n}$

Question

Prove the number of words $w$ of length $n$ on the alphabet $[n]$ such that $w_{n−i} > i$ and $w_{i} ≤ w_{i+1} + 1$ is ${\frac 1{n+1}}{2n\choose n}$

I am actually stuck on what is being asked. I thought I could build a relation, but the inequalities are throwing me off.