In the post, there is a formula called Weyl identity: \begin{align} \exp\left(-\,{k \over n}\,a_{n}\,z^{-n}\right) \exp\left({\ell \over n}\,a_{-n}\,w^{n}\right) = \exp\left({\ell \over n}\,a_{-n}\,w^{n}\right) \exp\left(-\,{k \over n}\,a_{-n}\,z^{-n}\right) \exp\left({k\ell \over n}\left[w \over z\right]^{n}\right) \end{align} Here $\left[a_n, a_m\right] = n\,\delta_{n\ +\ m,\,0\,\,}$.
I tried to verify this identity by expanding both sides of the identity. But I am not able to verify the identity. Any help will be greatly appreciated!
Look likes your version of Weyl's identity has a typo in the sign of one of its exponents.
If $[A,B] = c$ where $c$ is a scalar, then $$\frac{d}{dt} e^{tA} B e^{-tA} = e^{tA} [A,B] e^{-tA} = e^{tA} c e^{-tA} = c$$ This leads to $e^{tA} B e^{-tA} = B + ct$. In particular, $e^A B e^{-A} = B + c$. As a result, $$e^A e^B e^{-A} = e^{e^A B e^{-A}} = e^{B + c} = e^B e^c \quad\implies\quad e^A e^B = e^B e^A e^c. $$ Substitute $A$ by $-\frac{k}{n}a_n z^{-n}$, $B$ by $\frac{\ell}{n} a_{-n} w^n$ and notice
$$[ A, B ] = \left[ -\frac{k}{n}a_n z^{-n}, \frac{\ell}{n} a_{-n} w^n\right] = -\frac{k}{n}\frac{\ell}{n} \left(\frac{w}{z}\right)^n [ a_n, a_{-n} ] = -\frac{k\ell}{n}\left(\frac{w}{z}\right)^n $$ is a scalar, we find $$e^{-\frac{k}{n}a_n z^{-n}} e^{\frac{\ell}{n} a_{-n} w^n} = e^{\frac{\ell}{n} a_{-n} w^n} e^{-\frac{k}{n}a_n z^{-n}} e^{\color{red}{-}\frac{k\ell}{n}\left(\frac{w}{z}\right)^n}$$
Up to a sign in the exponents of last factor, this is the identity you have.