Binary heap: prove that number of nodes of height h is not bigger than $\lceil \frac{n}{2^{h+1}} \rceil$
$n$ element tree has depth $\lceil{\log_2n}\rceil-1$. If $h$ is the height of nodes in any $n$ element heap from leaves then the nodes are situated at the depth of $\lceil{\log_2n}\rceil-1-h$ from the root. 

Now, at most $2^{\lceil{\log_2n}\rceil-1-h} = \lceil{\frac{n}{2^{h+1}}}\rceil$ nodes/leaves are possible with $\lceil{\log_2n}\rceil-1-h$ depth in a heap.

Note that depth is the number of edges from the root &amp; height is the number of edges from the highest level in a heap.