Prove using induction that $2^n < \binom{2n}{n} < 4^n$ for $n \geq 2$ 
Trying to prove that, for $n\geq2$, $2^n < \binom{2n}{n} < 4^n$.

Inductive hypothesis: Assume $P(k)$ is true: 
\begin{align}
2^k < \binom{2k}{k} < 4^n \\\\
\end{align}
Show $P(k+1)$
\begin{align}
2^{k+1} < \binom{2k+2}{k+1} < 4^{k+1} \\\\
\end{align}
Not sure what's the strategy here... since it's a compound inequality I think it make sense to break the two apart and prove them separately?
Start with the inductive hypothesis
\begin{align}
2^k < \binom{2k}{k} < 4^n \\\\
\end{align}
Trying to prove the first inequality...
\begin{align}
2^k < \binom{2k}{k} \\\\
(2)2^k < (2)\binom{2k}{k} \\\\
2^{k+1} < (2)\binom{2k}{k} 
\end{align}
Stuck here now... not sure if I see what to do to the equation to get to 
\begin{align}
2^{k+1} < \binom{2k+2}{k+1} 
\end{align}
 A: For the induction step on the first inequality, you wish to show that if 
$$2^k < \binom{2k}{k}$$
then 
$$2^{k + 1} < \binom{2k + 2}{k + 1}$$
Using the definition of the binomial coefficient yields
\begin{align*}
\binom{2k + 2}{k + 1} & = \frac{(2k + 2)!}{(k + 1)!(k + 1)!}\\
                      & = \frac{(2k + 2)(2k + 1)(2k)!}{(k + 1)k!(k + 1)k!}\\
                      & = \frac{2(k + 1)(2k + 1)(2k)!}{(k + 1)k!(k + 1)k!}\\
                      & = 2 \cdot \frac{2k + 1}{k + 1} \cdot \frac{(2k)!}{k!k!}\\
                      & = 2 \cdot \frac{2k + 1}{k + 1}\binom{2k}{k}\\
                      & > 2\binom{2k}{k}\\
                      & > 2 \cdot 2^k & \text{by the induction hypothesis}\\
                      & = 2^{k + 1}
\end{align*}  
A: You can use the identity ${2k+2\choose k+1}={2k+1\choose k+1}+{2k+1\choose k}={2k\choose k+1}+2{2k\choose k}+{2k\choose k-1}=2{2k\choose k}+2{2k\choose k-1}>2{2k\choose k}$. 
For the other side, $2{2k\choose k}+2{2k\choose k-1}\leq 4{2k\choose k}$
