Calculate Laurent series for $1/ \sin(z)$

How can calculate Laurent series for

$$f(z)=1/ \sin(z) $$ ??

I searched for it and found only the final result, is there a simple way to explain it ?

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Using the series for $\sin(z)$ and the formula for products of power series, we can get

Using the formula for products of power series

As given in the Wikipedia article linked above,
\left(\sum_{k=0}^\infty a_kz^k\right)\left(\sum_{k=0}^\infty b_kz^k\right)
=\sum_{k=0}^\infty c_kz^k\tag{1}
1&\text{for }k=0\\
\frac{(-1)^j}{(2j+1)!}&\text{for }k=2j\\
0&\text{for }k=2j+1
Using $(2)$, $(3)$, and $(4)$, we can iteratively compute $b_k$.

For example, to compute the coefficient of $z^8$:
Thus, $b_8=\dfrac{127}{604800}$.

From the generating function of the Bernoulli polynomials:
$${\frac {t{{\rm e}^{xt}}}{{{\rm e}^{t}}-1}}=\sum _{n=0}^{\infty }{
\frac {{\it B}_{n}\left(x\right) {t}^{n}}{n!}} \tag{1}$$
and from:
$$ \frac{1}{\sin(z)}={\frac {-2i{{\rm e}^{-iz
}}}{{{\rm e}^{-2iz}}-1}} \tag{2}$$
we set $$t=-2iz,\quad x=\frac{1}{2} \tag{3}$$
in $(1)$ to get:
$${\frac {-2iz{{\rm e}^{-iz}}}{{{\rm e}^{-2iz}}-1}}=\sum _{n=0}^{
\infty }{\frac {{\it B_n} \left( \frac{1}{2} \right) \left( -2\,iz
\right) ^{n}}{n!}} \tag{4}$$
and we then use the known value that relates the polynomial evaluated at $\frac{1}{2}$, to the Bernoulli number:
$${\it B_n} \left( \frac{1}{2} \right) = \left( {2}^{1-n}-1 \right) {
\it B_n} \tag{5}$$
together with the fact that:
$${\it B}_{2n-1} =
\cases{1/2&$n=1$\cr 0&otherwise\cr} \tag{6}$$
and divide both sides of $(4)$ by $z$ to get:
\frac{1}{\sin(z)}&=\sum _{n=0}^{
\infty }{2{\it B}_{2n}\frac { (-1)^n\left( 1-2^{2n-1} \right) z^{2n-1}}{(2n)!}}\\
&=\frac{1}{z}+\frac{1}{6}\,z+{\frac {7
}{360}}\,{z}^{3}+{\frac {31}{15120}}\,{z}^{5}+…

Idea: $1=\sin z {1\over\sin z} =$ (known Taylor series)(unknown Laurent series).

EDIT: Solution here.