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.

\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}
