Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$ 
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$

Tried induction. Not sure where my mistake is, but what I did doesn't seem to make sense:
Let $n = 1.$ Then $1 + a + a^2 + \ldots +1 < \frac {1 + a + a^2 + \ldots +1 + a}{2} = \frac{2 + 2a + a^2+ \ldots}{2} = 1 + a + \frac{a^2}{2} + \ldots$
Then I did this below, but it's unclear if the difference is positive:
$\frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1} - \frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} = \frac {n + na +na^2+ \ldots + na^{n - 1} + na^n - n - 1 - an - a- na^2 - a^2 - \ldots -na^{n - 1} - a^{n - 1}}{n(n + 1)} = \frac{n(1 + a + a^2+ \ldots + a^{n - 1} + a^n - 1 -a - a^2 -a^{n - 1}) - 1 - a - a^2 - \ldots - a^{n - 1}}{n(n+1)} $
$ = \frac{- 1 + a(n(a^{n - 1}) - 1 - a - \ldots - a^{n - 2})}{n(n+1)}$
What can I try now? 
 A: Multiply both sides on $n(n-1)$  then you can see:
$$na^{n} > \sum_{i=0}^{n}a^{i}$$
Right side is geometric progression:
$$na^{n}>\frac{a^{n}-1}{a-1}$$
Now you can continue by yourself.
A: You have a more general result:

If $\{a_n \}$ is strictly increasing then also the sequence of the avarages over the first $N$ terms is strictly increasing. 

In fact we have
$$
S_N=\frac 1 N \sum_{k=1}^N a_k<\frac 1 N \sum_{k=1}^N a_N=a_N<a_{N+1}
$$
and therefore
$$
S_{N+1}=\frac {\sum_{k=1}^N a_k+ {a_{N+1}}} {N+1}=\frac{N S_N+a_{N+1}}{N+1}>
\frac{N S_N+S_N}{N+1}=S_N
$$
A: $$\frac{1+a+\ldots+a^{n-1}}n<\frac{1+a+\ldots+a^n}{n+1}\iff$$
$$\frac1n\,\frac{1-a^n}{1-a}<\frac1{n+1}\,\frac{1-a^{n+1}}{1-a}\iff$$
$$(n+1)(1-a^n)\stackrel{\text{since}\;1-a<0}>n(1-a^{n+1})\iff-na^n+1-a^n>-na^{n+1}\iff$$
$$n(a^{n+1}-a^n)>a^n-1\iff na^n(a-1)>(a-1)(a^{n-1}+a^{n-2}+\ldots+a+1)\iff$$
$$na^n>1+a+\ldots+a^{n-1}$$
and the last inequality is trivial since $\;a^n>a^k\;,\;\;\forall\,k=0,1,2,...,n-1\;$
A: Another solution:
Your inequality is equivalent to $\frac{a^n-1}{n}<\frac{a^{n+1}-1}{n+1}$. But this is $\int_1^a x^{n-1}dx< \int_1^a x^n dx$, which is true because $x^{n-1} < x^n, \forall x \in (1, a]$.
