Show that there are at most $\lceil n/2^{h+1} \rceil$ nodes of height $h$ in any $n$-element heap.

First, let's observe that the number of leaves in a heap is $\lceil n/2 \rceil$ (exercise 6.1-7). Let's prove it by induction on $h$.

**Base**: $h = 0$. The number of leaves is $\lceil n/2 \rceil = \lceil n/2^{0+1} \rceil$.

**Step**: Let's assume it holds for nodes of height $h-1$. Let's take a tree
and remove all it's leaves. We get a new tree with $n - \lceil n/2 \rceil =
\lfloor n/2 \rfloor$ elements. Note that the nodes with height $h$ in the old
tree have height $h-1$ in the new one.

We will calculate the number of such nodes in the new tree. By the inductive assumption we have that $T$, the number of nodes with height $h-1$ in the new tree, is:

$$ T = \bigg\lceil \lfloor n/2 \rfloor / 2^{h-1+1} \bigg\rceil < \bigg\lceil (n/2)/2^h \bigg\rceil = \bigg\lceil \frac{n}{2^{h+1}} \bigg\rceil $$

As mentioned, this is also the number of nodes with height $h$ in the old tree.