I think you are looking at the Bell numbers. Henry already succinctly explained how to get this result by finding the recurrence of the Stirling numbers of the second kind (which count partitions of an $n$-set into $k$ unlabelled blocks), but I will add a bit more detail. When going from an $N-1$ stack to an $N$ stack, one of two situations arises: either the new disk is larger than any previous one, or at least one previous disk is at least as large. Designate by $k$ the size of the largest disk, then in the first situation the $N-1$ stack has $k-1$ as its largest size (and there is only one way to extend it to a stack with $k$ as largest size), while in the second situation the $N-1$ stack has $k$ as its largest size, and it can be extended by any disk of size from $1$ to $k$. Denoting by $S(N,k)$ the number of stacks of height $N$ with $k$ the size of the largest disk, we get the recurrence relation
S(N,k) = S(N-1,k-1) + kS(N-1,k)
that Henry gave. You can recognise this (together with $S(N,1)=1$ and $S(1,k)=0$ for $k>1$) as the relation defining the Stirling numbers of the second kind, or in case you "forgot" that, you can compute an initial part of the (triangular) array of numbers, and recognise them, for instance by looking up in the OEIS.
Also let me clarify how you can relate your stacks to such paritions, with $k$ variable (so as to get the sum of the Stirling numbers over all $k$, giving the Bell number).
A stack partitions the set of the $N$ positions of disks (say $1$ for the bottom, up to $N$ for the top) into blocks of positions which hold disks of the same size $i$, with $i$ varying from $1$ to some maximal size $k$. To recover the stack from this partition (of which the parts are unlabelled), just label with $1$ the block that contains the bottom position $1$, then by $2$ the block with the lowest remaining position (i.e., not already in the first block), if any, then by $3$ the block with the lowest remaining position (i.e., not already in the first two blocks), if any, and so forth until labelling the last block $k$. Then put a disk of size $i$ into each position of the block labelled $i$, and that for $i=1,\ldots,k$. Clearly this satisfies your constraint, and establishes a bijection between stacks and partitions into blocks.