Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.

First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 1\cdot 2\cdot 2\cdot 2\cdot 3\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot 2^n\cdot n!}{(n!)^2}=2^n\cdot\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}.$
For $n=0$ we have:
$\frac{(2*0)!}{(0!)^2}=1=2^0.$
Now to our inductive step $n+1$:
\begin{align*}
& \frac{(2(n+1))!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(2n+2)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{1\cdot 2\cdot 3\cdot...\cdot 2n+1 \cdot 2n+2}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n \cdot 2n+2)}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot 2^{n+1} \cdot (n+1)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & 2^{n+1}\cdot \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)}{((n+1)!)}\geq2^{n+1}\\
\end{align*}

Is this proof correct/sufficient? Was induction even needed here or should I have argued instead that $\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}>0$? Is this step necessary even for the induction?

Another approach: Consider a line of $2n$ people, and we want to put them in two teams of $n$ people say $\color{red}{red}$ and $\color{blue}{blue}$ then what if you restrict to the first $n$ people that you put in the line, they can be whatever because regardless of the choice at most you picked $n$ of them in one team and so $2^n\leq \binom{2n}{n}.$
Example: Say $n = 2,$ then the possibilities are:
$$\underbrace{\color{red}{**}}_{\text{first 2}}\color{blue}{**},\hspace{1mm} \underbrace{\color{red}{*}\color{blue}{*}}_{\text{first 2}}\color{blue}{*}\color{red}{*},\hspace{1mm}\underbrace{\color{red}{*}\color{blue}{*}}_{\text{first 2}}\color{red}{*}\color{blue}{*},\hspace{1mm}\underbrace{\color{blue}{*}\color{red}{*}}_{\text{first 2}}\color{red}{*}\color{blue}{*},\hspace{1mm}\underbrace{\color{blue}{*}\color{red}{*}}_{\text{first 2}}\color{blue}{*}\color{red}{*},\hspace{1mm}\underbrace{\color{blue}{*}\color{blue}{*}}_{\text{first 2}}\color{red}{*}\color{red}{*},\hspace{1mm}.$$
Notice that restricting to the first $2$ we get all possibilities of red and blue.
