# Induction inequality related with $e$ number.

I'm stuck with induction problem. Let $$0\leq a, prove the next inequality for all $$n\in\mathbb{N}$$.

$$\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n$$I need this inequality because with this, we can deduce that the sequence $$\left(1+\frac{1}{n} \right)^{n}$$ converges.

The base, for $$n=1$$ it's trivial because $$\displaystyle\frac{b^2-a^2}{b-a}=b+a<2b$$ because by hypothesis $$a. Next suposse that for $$n=k$$ the inequality holds. If we take the inequality then $$\frac{b^{k+1}-a^{k+1}}{b-a}<(k+1)b^k$$ and multiplying for $$b$$ in both sides we obtain $$\frac{b^{k+2}-ba^{k+1}}{b-a}<(k+1)b^{k+1}<(k+2)b^{k+1}$$but from here I don't know how to obtain the result. Any hint? Thanks for your help.

Factor out $$\frac{b^{n+1}}{b}=b^n$$ and get
$$b^n \left(\frac{1-\left(\frac{a}{b}\right)^{n+1}}{1-\frac{a}{b}}\right)=b^n\sum_{k=0}^n\left(\frac{a}{b}\right)^k$$
by the geometric sum formula. Then since $$0\leq a we have for $$n>0$$
$$b^n\sum_{k=0}^n\left(\frac{a}{b}\right)^k < b^n\sum_{k=0}^n 1 = (n+1)b^n$$