Path to Combinatorics Problem [$1.11$] 
Problem: Let n be a positive integer greater than four and let $P_1P_2P_3..P_n$ be a convex n-sided polygon. Zachary wants to draw $n-3$ diagonals that partition the region enclosed by the polygon into $n-2$ triangular regions and that may intersect only at vertices of the polygon. In addition, he wants each triangular region to have at least one side that is a side of the polygon. In how many different ways can Zachary do this?

My attempt: I began with the case $n=5.$ I observed that the number of ways in which $3$ triangular regions can be constructed is $5$. I then went on to consider a hexagon and observed that the number of ways to partition was $(6-1)*5.$ Furthermore, I deduced for $n$ sided polygons we must have $(n-1)(n-2)...5*5$ ways to partition the polygon. I would like to know whether this answer is correct and if not, then a proof or a sketch would be much appreciated. 

I have found that the number is $n \times 2^{n-5}$. I begin: I choose a vertice $P_i$ of the polygon (the number of ways is $n$). I draw a triangular region $P_iP_{i+1} P_{i-1}$. Then, I have two choices: draw $P_{i+1}P_{i+2}P_{i-1}$ or $P_{i-1}P_{i-2}P_{i+1}$,...
If I just draw $P_aP_bP_c$ (with $b>c$), I have two choices: $P_bP_{b+1}P_c$ or $P_cP_{c-1}P_b$. 
