Proving $a^n - b^n \le na^{n - 1}(a-b)$ with induction The question is to prove $a^n - b^n \le na^{n - 1}(a-b)$ with induction where $0 < b < a$ and $ \forall n \ge 1$, $n \in \mathbb{Z}$
I finished the different parts of the proof, including the predicate, base case, and inductive step, but I want to make sure that I'm not missing anything - especially in the inductive step. 
I'm also a little confused about moving from Step 1 to the rest of the steps because it seems that the inductive hypothesis applies to $b(a^n-b^n$, but I'm not sure how the other part of the expression, namely $a^n(a-b)$ is used.
Let $P(n) = a^n - b^n \le na^{n - 1}(a-b)$
Base case: $P(0) = a - b \le a - b$
Inductive step - need to prove $P(n + 1)$:


*

*$a^{n + 1} - b^{n + 1} = a^{n + 1} - a^nb + a^nb - b^{n + 1} = a^n(a-b) + b(a^n - b^n)$

*Because $0 < b < a$, $b(a^n-b^n) \le a(a^n - b^n)$

*Using the induction hypothesis, $a(a^n - b^n) \le a \cdot n \cdot a^{n - 1}(a-b)$

*$\rightarrow a(a^n - b^n) \le na^{n}(a-b) \le (n + 1)a^{n}(a-b)$

*$\rightarrow a(a^n - b^n) \le (n + 1)a^{n}(a-b)$


\begin{align}
&a^{n+1} - b^{n+1} \\
=\ &a^{n+1} - a^nb + a^nb - b^{n+1} \tag{your step $1$} \\
=\ &a^n(a - b) + \color{red}{b(a^n - b^n)} \\
\leq\ &a^n(a - b) + \color{red}{a(a^n - b^n)} \\
\leq\ &a^n (a - b) + \color{red}{a \cdot na^{n-1}(a-b)} \\
\leq \ &a^n(a - b) + \color{red}{n\cdot a^n(a-b)} \tag{your step $4$}\\
=\ &(1 + n)a^n(a - b)
\end{align}
