inequality - If $a_1, a_2, ..., a_n$ are n positive real numbers such that $a_1.a_2....a_n=1$, then show that $(1+a_1).(1+a_2) ... (1+a_n) ge 2^n$

My attempt:

Using Arithmetic Mean $\ge$ Geometric Mean for $a_1, a_2, ..., a_n$,
$$\frac{a_1+a_2+...+a_n}{n} \ge \sqrt[n]{a_1.a_2....a_n}$$
$$\Rightarrow a_1+a_2+...+a_n \ge n$$

Therefore, $$\frac{(1+a_1)+(1+a_2)+...+(1+a_n)}{n} = \frac{(1+1+... n\;times) + (a_1+a_2+...+a_n)}{n} = \frac{n + (a_1+a_2+...+a_n)}{n} \ge \frac{n+n}{n}$$

Or simply, $$\frac{(1+a_1)+(1+a_2)+...+(1+a_n)}{n} \ge 2$$

Now, Using Arithmetic Mean $\ge$ Geometric Mean for $(1+a_1), (1+a_2), ..., (1+a_n)$,
$$\frac{(1+a_1)+(1+a_2)+...+(1+a_n)}{n} \ge \sqrt[n]{(1+a_1).(1+a_2)....(1+a_n)}$$

But, $$\frac{(1+a_1)+(1+a_2)+...+(1+a_n)}{n} \ge \sqrt[n]{(1+a_1).(1+a_2)....(1+a_n)}\;\;\;\;and\;\;\;\;\frac{(1+a_1)+(1+a_2)+...+(1+a_n)}{n} \ge 2$$
don't necessarily imply that $$\sqrt[n]{(1+a_1).(1+a_2)....(1+a_n)} \ge 2$$

This is where I require help to proceed further.