The inequality $b^n - a^n < (b - a)nb^{n-1}$ I'm trying to figure out why $b^n - a^n < (b - a)nb^{n-1}$.
Using just algebra, we can calculate
$ (b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) $
$ = (b^n + b^{n-1}a + \ldots + b^{2}a^{n-2} + ba^{n-1}) - (b^{n-1}a + b^{n-2}a^2 + \ldots + ba^{n-1} + a^{n-1}) $
$ = b^n - a^n, $
but why is it necessarily true that $(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}$?
Note:  I am interested in an answer to that last question, rather than in another way to prove the general inequality in the title...

This inequality can fail if $a=b$ or $a$ and $b$ have differing signs: e.g. $b=1$, $a=-3$, and $n=3$. So let's assume that $a\not=b$ and $a,b\ge0$. Division yields
$$
\frac{b^n-a^n}{b-a}=\sum_{k=1}^nb^{n-k}a^{k-1}\tag{1}
$$
If $a<b$ then obviously, $\sum\limits_{k=1}^nb^{n-k}a^{k-1}=b^{n-1}\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}< nb^{n-1}$, thus $\dfrac{b^n-a^n}{b-a}< nb^{n-1}$ and because $b-a>0$,
$$
b^n-a^n<(b-a)nb^{n-1}\tag{2}
$$
If $a>b$ then obviously, $\sum\limits_{k=1}^nb^{n-k}a^{k-1}=b^{n-1}\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}> nb^{n-1}$, thus $\dfrac{b^n-a^n}{b-a}> nb^{n-1}$ and because $b-a<0$,
$$
b^n-a^n<(b-a)nb^{n-1}\tag{3}
$$
If both $a$ and $b$ are negative, the inequality holds for even $n$, and is reversed for odd $n$.
