TITLE: Evaulate a determinant involving factorials.
QUESTION [5 upvotes]: In a problem set given by a teacher, there is the following problem.

If $a_n = \frac{1}{n!}$, evaluate
  $$ D_n = \begin{vmatrix}
a_1 & a_0 & 0 & 0 & \cdots & 0 & 0 & 0\\
a_2 & a_1 & a_0 & 0 & \cdots & 0 & 0 & 0 \\
a_3 & a_2 & a_1 & a_0 & \cdots & 0 & 0 & 0 \\
a_4 & a_3 & a_2 & a_1 & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
a_{n-2} & a_{n-3} & a_{n-4} & a_{n-5} & \cdots & a_1 & a_0 & 0 \\
a_{n-1} & a_{n-2} & a_{n-3} & a_{n-4} & \cdots & a_2 & a_1 & a_0 \\
a_n & a_{n-1} & a_{n-2} & a_{n-3} & \cdots & a_3 & a_2 & a_1
\end{vmatrix}. $$

By working out the first few values of $n$, I figured out that the answer is probably $D_n=a_n=\frac{1}{n!}$, but I have trouble proving this. I tried with induction.

Case $n=1$ is obvious.
We assume that $D_n=a_n$ is true for all $n$ up to $k$. We only need to prove that the statement holds for $n=k$ now.
By using Laplace's formula for the first row (and being lucky that $a_0=1$), we get
  $$ \begin{eqnarray}
D_k & = & a_1D_{k-1} - a_0\Big(
  a_2D_{k-2} - a_0\big( \cdots
     a_0\left( a_{n-1}D_1 - a_0a_n \right)
  \cdots \big)
\Big) \\
& = & \frac1{1!}\frac1{(k-1)!} - \frac1{2!}\frac1{(k-2)!} + \cdots + (-1)^{k-2} \frac1{(k-1)!}\frac1{1!} + (-1)^{k-1}\frac1{k!}
.\end{eqnarray}$$
  For uneven $k$, things are obvious. The other case is where things quickly get out of my hand.
For $k=2l$ we get $$ D_{2l} = 2\sum_{j=1}^{l-1}\frac1{j!\cdot(2l-j)!} + (-1)^{l+1}\left(\frac{1}{l!}\right)^2. $$
Not only that I've got another two cases for even and uneven $l$ (and I've got a bad feeling this might go on for a while), I also have no idea how to approach the sum.

Is there an easier way? I doubt I'm on the right track here (at least for even $k$).

REPLY [2 votes]: Let's play around your formula for $D_k$:
$$D_k=\frac1{1!}\frac1{(k-1)!} - \frac1{2!}\frac1{(k-2)!} + \cdots + (-1)^{k-2} \frac1{(k-1)!}\frac1{1!} + (-1)^{k-1}\frac1{k!} = \sum_{j=1}^k \frac{(-1)^{j+1}}{(k-j)!j!}
$$
$$=-\frac{1}{k!}\sum_{j=1}^k  (-1)^{j }{k\choose j} = \frac{1}{k!}{k\choose 0}-\frac{1}{k!}\sum_{j=0}^k  (-1)^{j }{k\choose j}=\frac {1}{k!}- (1-1)^k= \frac {1}{k!}.$$