<s>[INST] Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) = \frac{b^{n + 1} - a^{n + 1}}{(b-a)}$
What can I do from here?
 [/INST] If you look at $b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n$, you have $n+1$ terms. We have
$$
a^n + a^n +\cdots + a^n < b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n < b^n + b^n + \cdots + b^n
$$
because $a^n < a^kb^l < b^n$ if $k+l = n$ (and $k, l$ are both positive). Use the definition of multiplication to rewrite $a^n + \cdots +a^n = (n+1)a^n$ and $b^n + \cdots +b^n = (n+1)b^n$, and you're done.
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