Let the result be $a_n$ and the $n$ vertices be $A_1$, $A_2$, $\cdots$, $A_n$.

If there is no diagonal drawn from $A_1$, then there are $a_{n-1}$ ways to to divide the $(n-1)$-polygon $A_2A_3\cdots A_{n}$ into triangles, plus one additional triangle $\triangle{A_1A_{2}A_n}$.

Otherwise, if there is at least one diagonal drawn from $A_1$, let $k$ be the smallest index where $A_1A_k$ is drawn. Then, $3\le k\le n-1$ and there must exist $\triangle{A_1A_2A_k}$. Meanwhile, the $(k-1)$-polygon $A_2A_3\cdots A_k$ has $a_{k-1}$ ways to divide into triangles and the $(n-k+2)$-polygon $A_1A_kA_{k+1}\cdots A_n$ has $a_{n-k+2}$ ways. Therefore, we have $$a_n=a_{n-1}+\sum_{k=3}^{n-1}a_{k-1}a_{n-k+2}=a_{n-1}+\underbrace{\sum_{j=2}^{n-2}a_ja_{n-j+1}}_{let\ j=k-1}$$

where we define $a_2=1$ (thinking about when the case when $k=3$ or $k=n-2$). It follows that $a_{n-1}=a_2a_{n-1}$. Therefore, the above recursion can be written as $$a_n=\sum_{j=2}^{n-1}a_ja_{n-j+1}$$

where $a_2=1$ and $a_3=1$. There are several ways to solve this sequence.

Using Bijection

This recursion essentially the same as that in # 4477 except the index differs by $2$. Therefore the answer is $$\boxed{\frac{1}{n-1}\binom{2(n-2)}{n-2}}$$

Using Generating Function

See # 4511.