Identity involving logarithms

Question

Here's a picture from the book,

I don't get why the expression $$ n^{\log_ba-\epsilon} \sum_{j=0}^{\log_bn-1} a^j \left(\frac{1}{b^j}\right)^{\log_ba-\epsilon} = \sum_{j=0}^{\log_ba-1} \frac{(ab^\epsilon)}{(b^{\log_ba})^j} \text{ ?}$$

Answer 1

They didn't show few simplification steps in between:
$$ a^j\left(\frac{1}{b^j}\right)^{\log_ba-\epsilon} = \dfrac{a^j}{b^{j\log_ba -j\epsilon}}=\dfrac{a^j}{b^{j\log_ba}\cdot b^{ -j\epsilon}}=\dfrac{a^jb^{j\epsilon }}{b^{j\log_ba}}=\frac{(ab^\epsilon)^j}{(b^{\log_ba})^j} $$