You ask why it is necessarily true that
$$(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}.$$ A quick answer is that it is not necessarily true. We cannot have strict inequality if $b=a$. And there are other issues. For example, if $n=1$, we always have equality. And the inequality sometimes reverses when one of $a$ or $b$ is negative.
We deal in detail with the cases where $a$ and $b$ are both $\ge 0$. Suppose first that $b>a$, and let $n>1$. Then we want to show that
$$b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1} < nb^{n-1}.$$
The term $b^{n-2}a$ is strictly less than $b^{n-1}$, and there is such a term, since $n>1$. The remaining terms (if any) are also strictly less than $b^{n-1}$. There is a total of $n$ terms, so their sum is strictly less than $nb^{n-1}$.
A similar argument deals with $b nb^{n-1}.$$
Since $b1$. All the other terms are $\ge b^{n-1}$. There is a total of $n$ terms, and the inequality follows.
We stop at this point. The work is incomplete, since we have not dealt with negative numbers. Let $n=3$, $b=0$, and $a=-1$. Then $(b-a)(b^2+ba+a^2)$ is positive, but $(b-a)(3)b^2$ is $0$, so the inequality does not hold. With some work, one can track down all the situations where the inequality does hold.