We assume that $\ln{(S_T/S_0)} \sim N(\mu T, \sigma^2 T)$. I have in suggested solution that $$\ln(S_T/S_0) = \mu T + \sigma \sqrt{T} Z$$ where $Z \sim N(0,1)$ or $S_T = S_0e^{\mu T + \sigma \sqrt{T}Z}$. How do we know that $\ln{(S_T/S_0)} = \mu T + \sigma \sqrt{T}Z$? Also, when $S_T$ is lognormally distributed as here, we have
$$E[S_T^\alpha] = E\left[(S_0e^{\mu T + \sigma \sqrt{T}Z})^\alpha\right] = S_0^\alpha E\left[e^{\alpha \mu T + \alpha σ \sqrt{T}Z}\right] = S_0^\alpha e^{\alpha \mu T + \alpha^2 \sigma^2 T/2}$$
How did $E\left[e^{\alpha \mu T + \alpha σ\sqrt{T}Z}\right] = e^{\alpha \mu T + \alpha^2 \sigma^2 T/2}$ ?
Generally if $X \sim N(μ, σ)$ then $$Z=\frac{X-μ}{σ}\sim N(0,1)$$ Consequently here (with $\ln{(S_T/S_0)}$ in place of $X$) $$Z=\frac{\ln{(S_T/S_0)-μΤ}}{σ\sqrt{T}}\sim N(0,1)$$ Making the right (elementary) transformations yields the first relations. Now, the expectation is calculated as follows $$E[S_T^\alpha] = S_0^\alpha e^{\alpha \mu T}E\left[e^{\alpha σ \sqrt{T}Z}\right]=S_0^\alpha e^{\alpha \mu T} MGF_Z(\alpha σ\sqrt{T})$$ where $MGF_Z$ denotes the moment generating function of $Z$. For the normal distribution $ N(μ,σ^2)$ the MGF is given by $$MGF(t)=e^{tμ+\frac12σ^2t^2}$$ and for the standard normal $Z\sim N(0,1)$ which you need here $$MGF_Z(t)=e^{\frac12t^2}$$ and the result follows by substituting $t=ασ\sqrt{T}$.