Number of possible heaps on $\{1,...,2^h-1\}$ Let $C_h$ be the number of possible heaps for the set of keys $\{1,...,2^h-1\}$. Determine a recurrence relation for $C_h$ via the substitution method and prove it.
Definition
A binary tree is ordered if the key at every node is smaller than the keys at its children.
An ordered binary tree of height $h$ is called a heap if the tree induced by all levels except the last is a complete binary tree and all leaves are &quot;as much to the left as possible&quot;.

So what I've come up with is this expression:
$$C_h=\displaystyle\prod_{n=1}^{h} (2^{i-1})!$$
The intuition behind it is that with every  increment by $1$ of $h$ the tree gains $2$ times more leaves than before and the possibilities of ordering these leaves can be determined via the factorial operation. Does that make sense?
The definition you give looks like the definition of a complete tree. With the restriction that nodes are in $[\![1, 2^h-1]\!]$, then it is also a perfect tree of height $h$.

Instead of looking at the leaves, you should look at the root of the tree: since the tree is perfect, it has two children of size $2^{h-1}-1$ (this is how $C_{h-1}$ appears). You just have to think of how to split $2^{h}-2$ keys between left and right (the key of the root is uniquely determined, since it is a heap).

I will edit this answer if it is not clear enough.