A stack is considered valid if it meets the conditions. Let $a_{n}$ be the number of valid stacks with $n$ disks (with radii $1, 2, \ldots, n$). We prove by induction that $a_{n}=(n+1)!$. For $n=1$, we can make two stacks: one with the transparent disk of radius 1 and one with the opaque disk of radius 1, so $a_{1}=2=2!$. Suppose now that for some $n \geq 1$ we have proven that $a_{n}=(n+1)!$. Consider a valid stack with $n+1$ disks. If we remove the disk with radius $n+1$, all remaining disks are still visible from above, so we are left with a valid stack of $n$ disks. Every valid stack with $n+1$ disks can thus be created by inserting the disk with radius $n+1$ into a valid stack of $n$ disks at a suitable position. In principle, there are $n+1$ positions where we can insert the disk with radius $n+1$: above the top disk, above the second disk, ..., above the bottom disk, and also below the bottom disk. The disk with radius $n+1$ is always visible, regardless of where we insert it. If we insert the disk with radius $n+1$ below the bottom disk, it can be either transparent or opaque; in both cases, the view of the other disks is not blocked. Thus, there are $2 a_{n}$ valid stacks where the disk with radius $n+1$ is at the bottom. However, if we insert an opaque disk with radius $n+1$ at any other position, it blocks the view of the disks below it. Therefore, we can only insert the transparent disk with radius $n+1$ at the other $n$ positions. Thus, there are $n a_{n}$ valid stacks where the disk with radius $n+1$ is not at the bottom. We find

$$a_{n+1}=2 a_{n}+n a_{n}=(n+2) a_{n}=(n+2)(n+1)!=(n+2)!$$

This completes the induction.