Can I do this to retrieve the Euler numbers from the power series of $\sec(x)$ I know that we can derive the general formula series for functions like $\tan(x)$ or $\sec(x)$ elegantly using the compact sigma notation. I am still very clumsy with manipulating with these compacts so instead I just divide 1 for $\cos(x)$ (since $\sec(x)=\frac{1}{\cos(x)}$) and I obtain the following power series:
$\sec(x)=\color{green}{1}+\dfrac{\color{green}{1}x^2}{2!}+\dfrac{\color{green}{5}x^4}{24}+\dfrac{\color{green}{61}x^6}{720}+\dfrac{277x^8}{8064}+\dfrac{50521x^{10}}{3628800}...$
We know that the compact and useful general formula for $\sec(x)$ is
$$\sum_{n=0}^{\infty}\dfrac{(-1)^{n}E_{2n}x^{2n}}{(2n)!}$$
The even Euler numbers are $E_0=1$, $E_2=1$, $E_4=5$, $E_6=61$, $E_8=1385$, $E_{10}=50521$, $E_{12}=2702765...$
So the colored number matches the Euler numbers
But as you can see, due to the long division, and using symbolab to sum up terms in between during the process of long division, the fifth term $\dfrac{277}{8064}x^8$ is obscured, and it doesn't show the eighth Euler number, which is $1385$.
So, in order to improve the visibility, I reason as followed:
The Taylor formula tells us that the 8th term of this power series of $\sec(x)$ must be $\dfrac{f^{8}(0)}{(8!)}x^8$, the denominator of $\dfrac{277}{8064}x^8$ must be a number $x$ that is multiplied with $8!$ to produce 8064, in other words, $8!x=8064$. I then divide $\frac{8064}{8!}$ and obtain $\frac{1}{5}$. This means multiply the numerator with $5$, you get $277\cdot{5}=1385$, the  $E_8$.
To confirm this, I simply use the compact formula to calculate the 8th power and I get $\dfrac{1385}{40320}$, simplify both for $5$, I get the simplified version made by symbolab.
To further confirm this, I repeat the process for the $12th$ term of this series, which is $\dfrac{540553}{95800320}$, take $\frac{95800320}{12!}$, I get $5$, multiply this to the numerator, and I obtain the 12th Euler number $2702765$
My question is, is this the right way to retrieve the original numbers after simplification made by symbolab? Why does this work?

You were right, to see why the method works you can compare each term with the series given by symbolab i.e.
$$ \dfrac{E_{2n}}{(2n)!} = \dfrac{a}{b} $$ now you are multiplying a number $c$ such that $bc = (2n)!$ and if you put this in the equation you can easily see that $E_{2n} = ac$
