Here's the claim again, for reference:
Claim: For all positive integers \(n \geq 2\), \(2^n n! < (2n)!\)
Proof by induction on n.
Base: When n=2, \(2^n n! = 4 \cdot 2 = 8\) and \((2n)! = 4! = 24\), so \(2^n n! < (2n)!\) is true for this value of n.
Induction Hypothesis: Suppose that \(2^n n! < (2n)!\) is true for \(n = 2, 3, \ldots, k\) where k is an integer \(\geq 2\).
Rest of induction step: We need to show that \(2^{k+1} (k+1)! < (2(k+1))!\)
Notice that \((2(k+1))! = (2k+2)!\).
By the inductive hypothesis, we know that \(2^{k}k! < (2k)!\). Therefore \begin{align*} (2k+2)! & = (2k+2) (2k+1) (2k)! \\ &= (2k+1) \left[2 (k+1)\right] (2k)! \\ & > (2k+1) \left[2 (k+1)\right] 2^k k! \\ & = (2k+1) 2^{k+1} (k+1)! \\ \end{align*}
Since \(k \geq 2\), \((2k+1) \geq 5 > 1\). So
\((2k+1) 2^{k+1} (k+1)! > 2^{k+1} (k+1)!\).
So therefore we have
\((2k+2)!> (2k+1) 2^{k+1} (k+1)! > 2^{k+1} (k+1)!\)
So \(2^{k+1} (k+1)! < (2k+2)!\), which is what we needed to show.