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?
 A: 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}
A: 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.
