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?

Finally I found a direct way to work this out, that is, to use Generating Functions.

From this article we get the following proposition:

Proposition. Consider the following infinite matrix with $1$s in the super diagonal.
$$
D=\left[\begin{array}{ccccc}
b_{0} & 1 & 0 & 0 & \cdot \\
b_{1} & c_{1} & 1 & 0 & . \\
b_{2} & c_{2} & c_{1} & 1 & . \\
b_{3} & c_{3} & c_{2} & c_{1} & . \\
b_{4} & c_{4} & c_{3} & c_{2} & . \\
\cdot & . & . & . & .
\end{array}\right] 
$$
Let $B(x)=\sum_{n=0}^{\infty} b_{n} x^{n},$ and $C(x)=\sum_{n=1}^{\infty} c_{n} x^{n}$ be the generating functions for the sequences $b_{0}, b_{1}, b_{2}, \ldots$ and $c_{1}, c_{2}, c_{3}, \ldots,$ respectively. If
$$
A(x)=\frac{B(x)}{1+C(x)}=\sum_{n=0}^{\infty} a_{n+1} x^{n}
$$
then $a_{n}=(-1)^{n-1} D_{n}$ and $1+x A(-x)$ is the generating function of $D_{n}$.

For the original problem, it is equivalent to calculate the determinant:
$$
D=\left| \begin{matrix}
 \dfrac{1}{2!}&  1&  0&  \cdots&  0\\
 \dfrac{1}{3!}&  \dfrac{1}{2!}&  1&  \cdots&  0\\
 \dfrac{1}{4!}&  \dfrac{1}{3!}&  \dfrac{1}{2!}&  \cdots&  0\\
 \vdots&  \vdots&  \vdots&  \ddots&  \vdots\\
 \dfrac{1}{\left( 2k+2 \right) !}&  \dfrac{1}{\left( 2k+1 \right) !}&  \dfrac{1}{\left( 2k \right) !}&  \cdots&  \dfrac{1}{2!}\\
\end{matrix} \right|
$$
We can see that $\displaystyle B(x)=\sum _{i=0}^{\infty } \frac{x^{i}}{(i+2)!}= \frac{-x+e^x-1}{x^2}$, $\displaystyle C(x)=\sum _{i=1}^{\infty } \frac{x^i}{(i+1)!} = \frac{-x+e^x-1}{x}$, so $A(x)=\dfrac{1}{x}+\dfrac{1}{1-e^x}$, and finally the generating function for $D_n$ is $D(x)=\dfrac{x}{e^x-1}+x$.
Notice the order $N$ of original matrix $D$ is always odd (since $N=(2k+2)-2+1=2k+3$), so the original claim is equivalent to when $n>4$ and $n$ is odd, $[x^n]D(x)=0$. It's easy to see that $D(x) -1 - x/2$ is an even function, which implies its series only contains terms of the form $x^{2k}$, thus we finished our proof.

I think this method really requires a good understanding of linear equations and Cramer's Law, along with the knowledge of generating functions.
