Prove the determinant is $0$ This is Problem 16.17 from the book Exercises in Algebra by A. I. Kostrikin.

Prove that
$$
\left|\begin{array}{ccccc}
\dfrac{1}{2 !} & \dfrac{1}{3 !} & \dfrac{1}{4 !} & \cdots & \dfrac{1}{(2 k+2) !} \\
1 & \dfrac{1}{2 !} & \dfrac{1}{3 !} & \cdots & \dfrac{1}{(2 k+1) !} \\
0 & 1 & \dfrac{1}{2 !} & \cdots & \dfrac{1}{(2 k) !} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \dfrac{1}{2 !}
\end{array}\right|=0, \quad k \in \mathbb{N}
$$

My Attempt: I tried to expand it by the first column, but it seemed to be more complicated when I did that. I also tried to add edges to the determinant(in the hope that it will be easier to calculate), but I still failed to work it out.
So, My Question is, how to calculate this determinant?

We want to prove that the $(n-1)\times(n-1)$ matrix
$$
A_n=\pmatrix{
\dfrac{1}{2!} &\dfrac{1}{3!} & \cdots &\cdots & \cdots &\dfrac{1}{n!} \\
1 &\dfrac{1}{2!} &\dfrac{1}{3!} &\cdots &\cdots &\dfrac{1}{(n-1)!} \\
0 & 1 & \ddots & \ddots & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddots&\ddots & \vdots\\
\vdots & \ddots & \ddots &1& \dfrac{1}{2!} & \dfrac{1}{3!}\\
0 & \cdots & \cdots & 0 & 1 & \dfrac{1}{2!}
}
$$
is singular when $n\ge4$ is even. This can be proved by mathematical induction on $n$. The base case $n=4$ can be verified directly. In the inductive step, note that for any positive integer $m$,
$$
\sum_{k=0}^m\frac{(-1)^k}{k!}\frac{1}{(m-k)!}=\frac{1}{m!}\sum_{k=0}^m(-1)^k\binom{m}{k}=\frac{1}{m!}(1-1)^m=0.
$$
Move the first and the last summands out of the sum, we obtain
$$
\frac{1}{m!}+\sum_{k=1}^{m-1}\frac{(-1)^k}{k!}\frac{1}{(m-k)!}=\frac{(-1)^{m+1}}{m!}.
$$
Denote the $i$-th row of $A_n$ by $a_i$. The previous identity means that
\begin{aligned}
u&:=a_1+\sum_{k=1}^{n-1}\frac{(-1)^k}{k!}a_{k+1}\\
&=\left(-\frac{1}{2!},\,\frac{1}{3!},\,-\frac{1}{4!},\,\frac{1}{5!},\,\ldots,\,-\frac{1}{(n-2)!},\,\frac{1}{(n-1)!},\,\frac{1}{(n-1)!}-\frac{1}{n!}\right).
\end{aligned}
(Note that the last entry of $u$ is not $-\frac{1}{n!}$, because the last column of $A_n$ ends with $\frac{1}{2!}$, not $1$.) In other words, by some appropriate elementary row operations, we can modify the first row of $A$ to $u$. Therefore, by the multilinearity of the determinant function, $\det(A_n)$ remains unchanged if we replace the first row of $A_n$ by
$$
v:=\frac12(a_1+u)
=\left(0,\,\frac{1}{3!},\,0,\,\frac{1}{5!},\,\ldots,\,0,\,\frac{1}{(n-1)!},\,\frac{1}{2\times(n-1)!}\right).
$$
Now suppose $n\ge6$ is even. By Laplace expansion along $v$ and by induction assumption, we get
\begin{aligned}
\det(A_n)
&=-\frac{1}{3!}\det(A_{n-2})-\frac{1}{5!}\det(A_{n-4})-\cdots-\frac{1}{(n-1)!}\det(A_2)+\frac{1}{2\times(n-1)!}\\
&=-\frac{1}{(n-1)!}\det(A_2)+\frac{1}{2\times(n-1)!}\\
&=-\frac{1}{(n-1)!}\frac{1}{2!}+\frac{1}{2\times(n-1)!}\\
&=0.
\end{aligned}
