This is equivalent to proving the number of partitions in which the largest two parts are of different sizes is $p(n-1)$.
Given a partition $P$ of size $p(n-1)$ we assign it to the following partition of $n$.
If the partition has its two largest sets with equal size send that partition to the following partition: order the two largest sets depending on the minimimum element and send the partition $P$ to the partition which is equal to $P$ in every sense except for the fact that the $n$ belongs to the set that was one of the largest sets in $P$ and had the minimum element among those sets.
If the partition has its two largest sets with size $P$ then send the partition of $n$ that is identical to $P$ only $n$ belongs to a singleton set.
This function is clearly injective because given a partition of $n$ we can easily recover from which partition of size $n-1$ it came (and it is unique).
On the other hand it is not surjective, this is because if a partition of $n$ comes from a partition of $n-1$ and satisfies that its largest set has the element $n$ then there is always a set that has exactly $1$ element less than the largest set. While there are many partitions of $n$ in which the largest two sets have different sizes which satisfy that the largest set contains $n$, but the second smallest set does not have $1$ element less than the largest set.
In conclusion we gave a function from the set of partitions of $n-1$ to the set of partitions of $n$ in which the two largest sets have different sizes that it injective but no surjective, this tells us there are strictly less partitions of $n-1$ than partitions of $n$ in which the two largest sets have different sizes.