If AM-GM holds for any $2^n$ positive real numbers, show that AM-GM holds for *any number* of positive real numbers. I was able to prove by induction on $n\in\mathbb{N}$ that for any $m:=2^n$ positive real numbers we have $\text{Geometric Average}\leq\text{Arithmetic Average}$,
I.e. $\sqrt[m]{\Pi_{i=1}^m a_i} = $$\sqrt[2^n]{\Pi_{i=1}^{2^n} a_i} \leq\frac{\Sigma_{i=1}^{2^n} a_i}{2^n}$$ =\frac{\Sigma_{i=1}^m a_i}{m}$ where $a_1,...,a_m\in\mathbb{R}_+$ are any $m=2^n$ positive real numbers.
Now I am trying to prove by using this fact, That for any $n\in\mathbb{N}$ positive real numbers we have $\text{Geometric Average} \leq \text{Arithmetic Average}$,
I.e. $\sqrt[n]{\Pi_{i=1}^n a_i} \leq\frac{\Sigma_{i=1}^n a_i}{n}$ where $a_1,...,a_n\in\mathbb{R}_+$ are any $n$ positive real numbers, But then I got stuck and was not able to proceed.

In other words, The problem can be stated as:

Suppose that we know  $\forall n\in\mathbb{N},\forall a_1,...,a_{2^n}\in\mathbb{R}_+, \sqrt[2^n]{\Pi_{i=1}^{2^n} a_i} \leq\frac{\Sigma_{i=1}^{2^n} a_i}{2^n}$
and prove by using this fact that  $\forall n\in\mathbb{N},\forall a_1,...,a_{n}\in\mathbb{R}_+, \sqrt[n]{\Pi_{i=1}^{n} a_i} \leq\frac{\Sigma_{i=1}^{n} a_i}{n}$

Thanks for any hint/help...
($\mathbb{R}_+$ - The set of positive real numbers)
 A: Assume that:

For any $k\in\mathbb{N}$ and for any $b_1,...,b_{2^k}\in\mathbb{R}_+$, we have that:
  \begin{align}
\frac{b_1 + \cdots + b_{2^k}}{2^k} \geq \sqrt[2^k]{b_1 \cdots b_{2^k}} \tag{$\star$}
\end{align}

We want to show that:

For any $n\in\mathbb{N}$ and for any $a_1,...,a_n\in\mathbb{R}_+$, we have that:
  $$
\frac{a_1 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 \cdots a_n}
$$

To this end, choose any $n\in\mathbb{N}$ and choose any $a_1,...,a_n\in\mathbb{R}_+$. Since the sequence $\{2^i\}_{i=1}^\infty$ is unbounded above, we know that there exists some $k \in \mathbb N$ such that $n \leq 2^k$. Now define:
$$
\mu = \frac{a_1 + \cdots + a_n}{n}
$$
and for each $j \in \{1, \ldots, 2^k\}$, define:
$$
b_j = \begin{cases}
a_j &\text{if } j \leq n \\
\mu &\text{otherwise}
\end{cases}
$$
Then observe that:
\begin{align*}
\mu
&= \frac{a_1 + \cdots + a_n}{n} \\
&= \frac{2^k(a_1 + \cdots + a_n)}{2^kn} \\
&= \frac{n(a_1 + \cdots + a_n) + (2^k - n)(a_1 + \cdots + a_n)}{2^k n} \\
&= \frac{(a_1 + \cdots + a_n) + (2^k - n)\mu}{2^k} \\
&= \frac{b_1 + \cdots + b_{2^k}}{2^k} \\
&\geq \sqrt[2^k]{b_1 \cdots b_{2^k}} &\text{by $(\star)$} \\
&= \sqrt[2^k]{(a_1 \cdots a_n)\mu^{2^k - n}}
\end{align*}
Raising both sides to the power of $2^k$, we obtain:
$$
\mu^{2^k} \geq (a_1 \cdots a_n)\mu^{2^k - n}
$$
Multiplying both sides by $\mu^{n - 2^k}$, we obtain:
$$
\mu^n \geq a_1 \cdots a_n
$$
Finally, taking $n^{\text{th}}$ roots of both sides yields:
$$
\frac{a_1 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 \cdots a_n}
$$
as desired. $~~\blacksquare$
