$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$ proof by induction 
Prove with mathematical induction that for every $n \in \Bbb N$ it holds: 
  $$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$$



*

*Basis (P(1)) 



$\frac21=2^1$



*Inductive step - If it holds for P(n), then it also holds for P(n+1)



$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1} \cdot \frac{2(n+1)} {2(n+1)-1} = 2^n \cdot \frac{2n+2} {2n+1}$$

But this is not true, right? Because there are more numbers between $\frac{2n}{2n-1}$ and $\frac{2n+2} {2n+1}$ or something like that? In numerator, a number will be for one bigger than the previous number, and that's why what I have written in inductive step isn't true because there's a missing $2n+1$ in the numerator.
Then I tried to write it like this.

$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1} \cdot \frac{2n+1}{?}\cdot \frac{2n+2} {2n+1}$$

But then I don't know what to put in the denominator because in denominator, each number is for two bigger than the previous one, so the number should be $2n+1$ but that doesn't make any sense to me anymore.
I don't know what to do with this problem. How can this be proven?

In the induction step, you assume
$$\frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1}=2^n$$
and wish to show that
$$\frac{(n+1)+1}1 \cdot \frac{(n+1)+2}3 \cdots \frac{2(n+1)}{2(n+1)-1}=2^{n+1}.$$
You have
\begin{align}\frac{(n+1)+1}1 \cdot \frac{(n+1)+2}3 \cdots \frac{2(n+1)}{2(n+1)-1}&=\frac{n+2}1 \cdot \frac{n+3}3 \cdots \frac{2n+2}{2n+1} \\&=\frac{1}{n+1} \cdot \frac{n+1}1 \cdot \frac{n+2}3 \cdots \frac{2n}{2n-1} \cdot \frac{2n+1}{2n+1} \cdot \frac{2n+2}{1} \\
&= \frac{1}{n+1} \cdot 2^n \cdot \underbrace{\frac{2n+1}{2n+1}}_{=1} \cdot (2n+2) \textrm{  (by the induction hypothesis)}\\
&= \frac{2n+2}{n+1} \cdot 2^n \\
&= 2 \cdot 2^n \\
&= 2^{n+1} \end{align}
