I stumbled into this problem and as I was trying to solve it I got stuck. I found an answer for it but did not understand one step of it.
Consider the classical Black-Scholes model. We define a contingent claim $X$, which pays$$X= \max(S_T^\alpha -K,0)$$ at maturity $T>0$, for some constant $\alpha>0$
Determine the arbitrage-free price $V_t^X$ of $X$ at any time $t\in[0,T]$
I didn't have any problem understanding the first part as it is just applying the solution for the Geometric Brownian SDE
$V_t^X=e^{rt}E_Q[e^{-rT}\max(S_T^\alpha -K,0 )|F_t]$
$V_t^X =e^{-r(T-t)}E_Q[\max(S_t^\alpha e^{\alpha(r-\frac{1}{2}\sigma^2)(T-t)+\alpha\sigma(W_T^Q-W_t^Q)}-K,0)|F_t]$
$V_t^X =e^{-r(T-t)}E_Q[\max(S_t^\alpha e^{\alpha(r-\frac{1}{2}\sigma^2)(T-t)+\alpha\sigma(W_T^Q-W_t^Q)+\frac{1}{2}\alpha^2\sigma^2(T-t)-\frac{1}{2}\alpha^2\sigma^2(T-t)}-K,0)|F_t]$
The following step is what I didn't understand
$V_t^X =e^{-r(T-t)} S_t^\alpha e^{\alpha(r-\sigma^2)(T-t)+\frac{1}{2}\alpha^2\sigma^2(T-t)}E_Q[\max(e^{\alpha\sigma(W_T^Q-W_t^Q)-\frac{1}{2}\alpha^2\sigma^2(T-t)}-\frac{K}{S_t^\alpha e^{\alpha(r-\sigma^2)(T-t)+\frac{1}{2}\alpha^2\sigma^2(T-t)}},0)|F_t]$
Why is it that the term $S_t^\alpha e^{\alpha(r-\frac{1}{2}\sigma^2)(T-t)}$ turns into
$S_t^\alpha e^{\alpha(r-\sigma^2)(T-t)}$ when I take it out of the expectation ?