This is just AM-GM: $\frac{1+a_k}{2}\geq\sqrt{a_k}$

This is such an old chestnut ... anything one tries will work! The obvious simplest way is by using $1+a_k \geq 2\sqrt{a_k}$ for all $1\leq k \leq n$. Then $\prod_{k=1}^n (1+a_k) \geq \prod_{k=1}^n 2\sqrt{a_k} = 2^n \sqrt{\prod_{k=1}^n a_k} = 2^n$.

Apply Holder's inequality: $ (1+a_{1})(1+a_{2})\cdots (1+a_{n})\ge (1+\sqrt[n]{a_1a_2a_3\ldots a_n})^n=(1+1)^n= 2^{n} $ The inequality is true if $a_1a_2a_3\ldots a_n \ge 1$ also.

We have $ (1+a_1)(1+a_2)\cdots (1+a_n)=$ $1+a_1+\cdots +a_n+a_1a_2+a_2a_3+$ $\cdots +a_{n-1}a_n+a_1a_2a_3+\cdots +a_{n-2}a_{n-1}a_n+\cdots +a_1a_2\cdots a_n $ and there are $ {n\choose 2} a_ia_j$, ${n\choose 3} a_ia_ja_k $ etc. Now $ a_1+a_2+\cdot d +a_n\ge$ $ n\sqrt[n]{a_1a_2\cdots a_n}=n$, $a_1a_2+\cdots +a_{(n-1)}a_n\ge\ { n\choose 2} + \cdots + a_1a_2\cdots a_n\ge {n\choose n} $. So $ LHS\ge\ 1+n+{n\choose 2}+{n\choose 3}+\cdots +{n\choose n}=(1+1)^n=2^n $.

APPLY AM GM TO ALL BRACKETS INDIVIDUALLY AND USE THE GIVEN CONDITION TO GET THE REQUIRED RHS

Very good and interesting solution for old and easy problem.Thank you @andrejilievski.

Notice that $1+a_k\geq 2\sqrt{a_k}$ by AM-GM. Then $(1+a_1)(1+a_2)\cdots (1+a_n)\geq 2^n\sqrt{a_1a_2\dots a_n}=2^n$. Equality occurs at $a_1=a_2=\cdots =a_n=1$.

By AM-GM on $1$ and $a_i$, $\tfrac{1+a_i}{2}\geq\sqrt{1\cdot a_i}\iff1+a_i\geq2\sqrt{a_i}$. Multiplying over $i=1,2,\dots,n$, $$\prod_{i=1}^n(1+a_i)\geq\prod_{i=1}^n2\sqrt{a_i}=2^n\sqrt{\prod_{i=1}^na_i}=2^n\sqrt{1}=2^n,$$as desired. Equality holds at $a_1=a_2=\dots=a_n=1$.

Nice, one. Note, that, for all $a_i, \frac{1+a_i}{2} \ge \sqrt{a_i} \implies 1+a_i \ge 2\sqrt{a_i},$(follows by AM-GM) now, if we multiply, over all $n,$ we have $2^n \cdot \sqrt{a_1\cdot a_2\cdot a_3\cdot ...... \cdot a_n}=2^n,$ using the fact $a_1\cdot a_2\cdot a_3\cdot ..... \cdot a_n=1.$

By Hölder's inequality, $(1+a_1)^{1/n}(1+a_2)^{1/n}\cdots(1+a_n)^{1/n}\geq (1\cdot1\cdots1)^{1/n}+(a_1a_2\cdots a_n)^{1/n}=2$. Raising both sides to the $n$th power yields the desired inequality.

By AM-GM, $1 + a_i \ge 2\sqrt{a_i}$. Taking the cyclic product, $$\prod_{i = 1}^n (1+ a_i) \ge 2^n \sqrt{\prod_{i = 1}^n a_i} = 2^n.\square$$