Is it true that $b^n-a^n < (b-a)nb^{n-1}$ when $0 < a< b$? A Real Analysis textbook says the identity
$$b^n-a^n = (b-a)(b^{n-1}+\cdots+a^{n-1})$$ yields the inequality
$$b^n-a^n < (b-a)nb^{n-1} \text{ when } 0 < a< b.$$
(Note that $n$ is a positive integer)
No matter how I look at it, the inequality seems to be wrong. Take for instance, the inequality does not hold for $n=1$ when one tries mathematical induction. It does not hold for other values of $n$ too. I guess there is something I am missing here and I will appreciate help.
 A: Observe that $n > 1$ for the assertion to be valid. Thus:  $b^{n-1-k}a^k< b^{n-1-k}b^k=b^{n-1}$. Letting $k$ runs from $0$ to $n-1$ and add them up: $b^{n-1}+b^{n-2}a+\cdots+a^{n-1} < nb^{n-1}$ which implies the inequality in question.
A: We basically need to show 
$$  b^{n-1}+\ldots+a^{n-1}<nb^{n-1}$$
Since $a<b$, then $a^{n-1}<b^{n-1}$. There are $n$ terms, and so the inequality holds only for when $n>1$.
Note that there is the same question here which uses a $\le$ sign, so i think it is a misprint in the text.
A: \begin{align}
b^n-a^n & = (b-a)(b^{n-1}+ b^{n-2}a + b^{n-3}a^2 + b^{n-4}a^3 + b^{n-5} a^4 +\cdots+a^{n-1}) \\[10pt]
& < (b-a)(b^{n-1} + b^{n-2} b + b^{n-3}b^2 + b^{n-4}b^3+ b^{n-5}b^4 + \cdots + b^{n-1}) \\[10pt]
& = (b-a)(b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + \cdots + b^{n-1}) \\[10pt]
& = (b-a) n b^{n-1}.
\end{align}
The only positive integer $n$ for which this does not work is $n=1,$ where the second factor has only one term, which is $1.$ And in that case it works if you say $\text{“}\le\text{''}$ instead of $\text{“}<\text{''}.$
\begin{align}
b^2-a^2 & = (b-a)(b+a) < (b-a)(b+b) & & = (b-a)2b. \\[10pt]
b^3-a^3 & = (b-a)(b^2 + ba + a^2) < (b-a)(b^2+b^2+b^2) & & = (b-a)3b^2. \\[10pt]
b^4 - a^4 & = (b-a)(b^3+b^2a+ba^2+a^3) \\
& < (b-a)(b^3+b^3+b^3+b^3) & & = (b-a)4b^3. \\[10pt]
b^5-a^5 & = (b-a)(b^4 + b^3a + b^2 a^2 + ba^3 + a^4) \\
& < (b-a)(b^4+b^4+b^4+b^4+b^4) & & = (b-a)5b^4. \\[10pt]
& \qquad\qquad\text{and so on.}
\end{align}
A: As $0 < a < b$ then $a^k < b^k$ and....
\begin{align}
b^n-a^n & = (b-a)(b^{n-1}+b^{n-2}a+\cdots+a^{n-2}b + a^{n-1}) \\
& =(b-a)\sum_{k=0}^{n-1} b^{n-k-}a^k \\
& < (b-a)\sum_{k= 0}^{n-1}b^{n-k-} b^k \\
& = (b-a)\sum_{k=0}^{n-1} b^{n-1} \\
& =(b-a)nb^{n-1}.
\end{align}
