How was this step done in order to get $n^\alpha$?

Question

In the following calculation, I don't understand how they got $n^\alpha$.

$Y = nB\left(\frac {(1-u)N}n \right)^{1-\alpha}$

Taking logs on both sides:

$y = \ln Y = \ln n^\alpha + \ln B + (1-\alpha)\ln[(1-u)N]$

Answer 1

$Y = nB\left(\frac {(1-u)N}n \right)^{1-\alpha}$

$\hphantom{Y} = n^{1-(1-\alpha)} B ((1-u)N)^{1-\alpha}$

$\hphantom{Y} = n^\alpha B ((1-u)N)^{1-\alpha}$

$$$$

$\therefore \ln Y = \ln n^\alpha + \ln B + (1-\alpha)\ln [(1-u)N]$