Prove that: $${\rm{E}}_{2n}=(-1)^n(2n)!\left|{\begin{array}{ccccccc} \frac{1}{2!} & 1 & 0 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{4!} & \frac{1}{2!} & 1 & 0 & \cdots & 0 & 0\\ \\ \frac{1}{6!} & \frac{1}{4!} & \frac{1}{2!} & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ \\ \frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \frac{1}{(2n-8)!} & \cdots & \frac{1}{2!} & 1\\\\ \frac{1}{(2n)!} &\frac{1}{(2n-2)!} & \frac{1}{(2n-4)!} & \frac{1}{(2n-6)!} & \cdots & \frac{1}{4!} & \frac{1}{2!}\end{array}}\right|$$
where ${\rm E}_{2n}$ are the Euler numbers with even index. (given also http://en.wikipedia.org/wiki/Euler_number )
I started the problem using the fact that: $$\sec x =\sum_{n=0}^{\infty}\frac{(-1)^n {\rm E}_{2n}x^{2n}}{(2n)!}, \;\;\; \left | x \right |<\frac{\pi}{2}$$
Hence: $$\sec x \cdot \cos x =1 \implies \sum_{n=0}^{\infty}\frac{(-1)^n{\rm E}_{2n}x^{2n}}{\left ( 2n \right )!}\cdot \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=1$$
Now I have to count , closely, in order to get what I want. Now I got stuck.. How can I proceed further? I cannot seem to combine the final result to get I want.
Thank you.
By expanding these determinants with the Laplace rule, you can simply check that they satisfy the same recursion satisfied by the Euler numbers, and that the identity holds for $n=0$ and $n=1$.
See also this other question (line $(2)$ in robjohn's answer).