Find the Black-Scholes price of an option paying $$(S_T^{\alpha} - K)_{+}$$ at time $T$.
Solution - The forward price is given by
$$F_T(t) = e^{r(T-t)}S_t$$
So,
$$F_T(0) = e^{rT}S_0$$
and
$$F_T(T) = S_T = F_T(0)e^{-\frac{1}{2}\sigma^2 T + \sigma\sqrt{T}N(0,1)}$$
So,
\begin{align*} F_T(T)^{\alpha} &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \sigma\alpha\sqrt{T}N(0,1)}\\ &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}e^{- \frac{\sigma^2 \alpha^2}{2} + \sigma\alpha\sqrt{T}N(0,1)} \end{align*}
Then use the Black formula for a call option with forward price
$$F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}$$
and volatility $\alpha\sigma$.
Question:
I do not understand why Joshi splits up the exponential in this part of the solution:
\begin{align*} F_T(T)^{\alpha} &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \sigma\alpha\sqrt{T}N(0,1)}\\ &= F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}e^{- \frac{\sigma^2 \alpha^2}{2} + \sigma\alpha\sqrt{T}N(0,1)} \end{align*}
I do not understand the logic of then concluding that we use the Black formula for a call option with forward price
$$F_T(0)^{\alpha}e^{-\frac{1}{2}\sigma^2 T \alpha + \frac{\sigma^2 \alpha^2}{2}}$$
and volatility $\alpha \sigma$.
Lastly, in exercise 21 we are asked to price the put $(K - S_T^{\alpha})_{+}$. The steps are exactly the same exact the volatility term is $\alpha\sigma \sqrt{T}$, which does not make sense to me. Any suggestions on these points are greatly appreciated.