I suppose we can see the edge of transparent disk (otherwise the problem doesn't make much sense). For each subset $S$ of $[n]= \{ 1,2,...,n\}$, if we know that the stack will use opaque disk of radii $s$ if and only if $s\in S$. We get that the order of those opaque disks is fixed. If $S$ is not an empty set, let $S=\{ s_1,s_2,...,s_k\}$ where $s_1<s_2<...<s_k$ and $k=|S|$. For convenient, let $s_0=0$ and $s_{k+1}=n+1$. We get that we must arrange disks of radii in the interval $(s_i,s_{i+1})$ so that all such disks are above of the disk with radius $s_{i+1}$. Inductively put the disk with radius not appeared, from smallest to largest, we get that there are $$\prod_{i=0}^{k}{\prod_{j=s_i+1}^{s_{i+1}-1}{j}}=\frac{n!}{\prod_{i=1}^{k}{s_i}}$$distinct stacks (here we denote $\prod_{k=a}^{b}{k}=1$ for all positive integers $a>b$). Note that if $S$ is an empty set, there are $n!$ ways of stacking. Hence, we want to calculate $n!+\sum_{S\subseteq [n],S\neq \emptyset}{\frac{n!}{\prod_{s\in S}{s}}}$. It's equal to $n!\prod_{i=1}^{n}{\Big( \frac{1}{1}+\frac{1}{i}\Big) }=(n+1)!$, done.

I found the answer is - $$\sum_{k=0}^{n} (n-k)! \times {(n-k+1)k \choose k} $$Is there any wrong? If not how can I manipulate it, Please help! $\color{green} \textbf{Explanation :}$ We will set $\mathbb{O}= (o_1,o_2,...o_k)$ be the set of opaque disks and they must be fixed from above to down by $o_1<o_2...<o_k$ Firstly, We assume $(n-k)$ places for remaining $(n-k)$ transparent disks. they can be placed in these places by $(n-k)!$ ways. Now assume there are $(n-k+1)$ spaces at the front and end and between the $(n-k)$ opaque disks, also let each of these $(n-k+1)$ spaces are divided into $k$ sub-spaces. We wish to place the $k$ opaque disks in these $(n-k+1)k$ sub-spaces in ${(n-k+1)k \choose k}$ ways, and the remaining spaces will be then removed