A supposedly straight forward proof The authors of this paper claim that equation (1) is equivalent to equation (2). I really don't see how this is possible. There even go further to say, "it is easy to check". Can anyone help?.

 A: First note that:
$$
e^{\ln(x)} = x
$$
and
$$
e^{-\ln(x)} = \frac{1}{x}
$$
$$\prod_{i=1}^G (X_g/\nu)^{1/G} = Z_{pool}
$$
$$
\begin{align*}
\overline{\ln(X_g)} &= \sum_g \ln(X_g)/G \\
&= \ln(\prod_g X_g)/G\\
&= \ln(\prod_g X_g^{1/G})\\
&= \ln(\prod (Z_g\nu)^{1/G})\\
\end{align*}
$$
$$
\begin{align*}
e^{-\overline{\ln(X_g)}} &= \frac{1}{\prod_g (Z_g\nu)^{1/G}}\\
&=\frac{1}{Z_{pool}\nu^{G/G}}\\
&=\frac{1}{Z_{pool}\nu} 
\end{align*}
$$
$$
\begin{align*}
\exp((1-\alpha) \times  (\ln(X_g) - \overline{\ln(X_g)}) &= \big(e^{(\ln(X_g) - \overline{\ln(X_g))}} \big)^{(1-\alpha)}\\
&= \exp(\ln(X_g))^{(1-\alpha)} \times \exp(-\overline{\ln(X_g)})^{(1-\alpha)}\\
&= (X_g)^{(1-\alpha)} \times \big(\frac{1}{Z_{pool}\nu} \big)^{(1-\alpha)}\\
&= (X_g/\nu)^{(1-\alpha)}\times \big(\frac{1}{Z_{pool}} \big)^{(1-\alpha)}\\
&= Z_g^{(1-\alpha)}\big(\frac{1}{Z_{pool}} \big)^{(1-\alpha)}
\end{align*}
$$
Thus,
$$
\begin{align*}
\prod_{i=1}^G (X_g/\nu)^{1/G}  \times \exp((1-\alpha) \times  (\ln(X_g) - \overline{\ln(X_g)}) &= Z_{pool} \times Z_g^{(1-\alpha)}\big(\frac{1}{Z_{pool}} \big)^{(1-\alpha)}\\
&= Z_{pool}^{\alpha} (Z_g)^{(1-\alpha)}
\end{align*}
$$
