For $0For $0<a<b$, how does $b^n-a^n=(b-a)(b^{n-1}+ab^{b-2}+...+a^{n-1})\implies b^n-a^n<(b-a)nb^{n-1}$?
Clearly, $$\frac{b^n-a^n}{b-a}=b^{n-1}+ab^{n-2}+...+a^{n-1}$$ so I would need to see that $$b^{n-1}+ab^{n-2}+...+a^{n-1}<nb^{n-1}$$
Dividing through by $b^{n-1}$:$$1+\frac{a}{b}+\frac{a^2}{b^2}+...+\frac{a^{n-1}}{b^{n-1}}<n$$
Why should I believe this last inequality even? 

$0 < a < b$ so $a^kb^{n-k-1} < b^{n-1}$.
So $b^{n-1} + ab^{n-2} + ...+ba^{n-2} + a^{n-1} < b^{n-1} + b^{n-1} + ...... + b^{n-1} + ^{n-1} = nb^{n-1}$.
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$1+\frac{a}{b}+\frac{a^2}{b^2}+...+\frac{a^{n-1}}{b^{n-1}}<n$ 
"Why should I believe this last inequality even? "
Because $\frac {a^k}{b^k} < 1$ so 
$1+\frac{a}{b}+\frac{a^2}{b^2}+...+\frac{a^{n-1}}{b^{n-1}} <1+1+1+......+1 = n$.
