Let's take the left child of the node indexed by ⌊𝑛/2⌋+1⌊n/2⌋+1.
LEFT(⌊𝑛/2⌋+1)
=2(⌊𝑛/2⌋+1)
>2(𝑛/2−1)+2
=𝑛−2+2
=n.
Since the index of the left child is larger than the number of elements in the heap, the node doesn't have childrens and thus is a leaf. The same goes for all nodes with larger indices.
Note that if we take element indexed by ⌊n/2⌋, it will not be a leaf. In case of even number of nodes, it will have a left child with index n and in the case of odd number of nodes, it will have a left child with index n−1 and a right child with index n.
This makes the number of leaves in a heap of size n equal to ⌈n/2⌉.