Permuting 15 books about 2 shelves, with at least one book on each shelf.

Breaking it into stages is the right idea, but the stages aren’t the two shelves. The first stage is putting all $15$ books onto the first shelf in some particular order. The second stage is deciding where to ‘cut’ that line of books to transfer the tail end to the second shelf. Can you finish it from there?

Added: There is a way to get the result by thinking first of how to split the books between the shelves, but it’s a bit messier. There are $\binom{15}k$ ways to decide which $d$ books to put on the first shelf. Once we’ve put them there, we can sort them in $k!$ ways, and independently we can sort the $n-k$ books on the second shelf in $(n-k)!$ ways. Since we have to have at least one book on each shelf, $k$ can range from $1$ through $14$, and the desired number is

$$\sum_{k=1}^{14}\binom{15}kk!(n-k)!=\sum_{k=1}^{14}\frac{15!}{k!(n-k)!}k!(n-k)!=\sum_{k=1}^{14} 15!=14\cdot15!\;.$$