Show this matrix is invertible Consider the matrix
\begin{equation}
S_n = 
  \begin{bmatrix}
   \frac{1}{2!} & \frac{1}{3!} & \cdots & \frac{1}{(n+1)!} \\
   \frac{1}{3!} & \frac{1}{4!} & \cdots & \frac{1}{(n+2)!} \\
   \vdots & \vdots & \ddots & \vdots \\
   \frac{1}{(n+1)!} & \frac{1}{(n+2)!} & \cdots & \frac{1}{2n!} \\
  \end{bmatrix}.
\end{equation}
I am interested to know if $S_n$ is invertible for all $n$. I do not need the inverse explicitly, knowing it exists is sufficient (ie. non-zero determinant for all $n$ is enough).
If this matrix has a special name, I would appreciate if someone could bring it to my attention.

For each $k\le n$, define a $k\times k$ matrix as follows:
$$
A_k=\begin{bmatrix}
\dfrac{(n+1)!}{(n-k+2)!} &\dfrac{(n+2)!}{(n-k+3)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+1)!}\\
\dfrac{(n+1)!}{(n-k+3)!} &\dfrac{(n+2)!}{(n-k+4)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+2)!}\\
\vdots &\vdots &\ddots &\vdots &\vdots\\
\dfrac{(n+1)!}{(n-1)!} &\dfrac{(n+2)!}{n!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+k-2)!}\\
\dfrac{(n+1)!}{n!} &\dfrac{(n+2)!}{(n+1)!} &\cdots &\cdots &\dfrac{(n+k)!}{(n+k-1)!}\\
1&1&\cdots&\cdots&1
\end{bmatrix}
$$
Since $A_n=S_n\operatorname{diag}\left((n+1)!,(n+2)!,\ldots,(2n)!\right)$, it suffices to show that $A_k$ is invertible for each $1\le k\le n$. We shall prove this by mathematical induction. The base case $k=1$ (with $A_1=[1]$) is trivial. For the inductive step, subtract every $j$-th column (with $j\ge2$) by the column on its left to obtain
$$
\left[\begin{array}{c|cccc}
\dfrac{(n+1)!}{(n-k+2)!} &\dfrac{(n+1)!}{(n-k+3)!}(k-1) &\cdots &\cdots &
\dfrac{(n+k-1)!}{(n+1)!}(k-1)\\
\dfrac{(n+1)!}{(n-k+3)!} & \dfrac{(n+1)!}{(n-k+4)!}(k-2) &\cdots &\cdots &\dfrac{(n+k-1)!}{(n+2)!}(k-2)\\
\vdots &\vdots &\ddots &\vdots &\vdots\\
\dfrac{(n+1)!}{(n-1)!} &\dfrac{(n+1)!}{n!}(2) &\cdots &\cdots &\dfrac{(n+k-1)!}{(n+k-2)!}(2)\\
n+1 &1 &\cdots &\cdots &1\\
\hline
1&0&\cdots&\cdots&0
\end{array}\right].
$$
Now we are done because the top right sub-block of this matrix is $\operatorname{diag}(k-1,k-2,\ldots,1)A_{k-1}$. This proof also shows that $|\det S_n|=\prod_{k=1}^n\frac{(k-1)!}{(n+k)!}$.
