Copy the steps below and fill in the blanks to prove that $n ! \gt 2^{n}$ for $n ≥ 4$. Note that we do not start with $n = 1$. Verify that the statement is false for $n \lt 4$.

1. Verify the statement is true for $n = 4$.

   Answer (a):

   $4! = 24 \gt 2^{4} = 16$ whereas $3! = 6 ≯ 2^{3} = 8$.

2. Assume the statement is true for $n = k$.

   Answer (b):

   Assume that $k! \gt 2^{k}$ for $k ≥ 4$.

3. Prove the statement is true for $n = k + 1$.

   Recall that: $\left(k + 1\right)! = \rule{.5cm}{0.15mm} \left(k + 1\right)$

   We assumed that $k ! \gt 2^{k}$, therefore $\left(k + 1\right)! = \rule{.5cm}{0.15mm} \left(k + 1\right) \gt \rule{.5cm}{0.15mm} \left(k + 1\right)$.

   Because $k ≥ 4$, $k + 1 \gt 2$ and hence ____ $(k + 1) \gt 2^{k} \cdot \rule{.5cm}{0.15mm}=2^\rule{.5cm}{0.15mm}$.
   Following the steps from above, we have shown that ____ $\gt$ ____.

   Answer (c):

   Prove the statement is true for $n = k + 1$.

   Recall that: $\left(k + 1\right)! = \color {red}{k!} \left(k + 1\right)$

   We assumed that $k ! \gt 2^{k}$, therefore $\left(k + 1\right)! = \color {red}{k!} \left(k + 1\right) \gt \color{red}{2^k} \left(k + 1\right)$.

   Because $k ≥ 4$, $k + 1 \gt 2$ and hence $\color{red}{2^k} (k + 1) \gt 2^{k} \cdot \color{red}2=2^{\color{red}{k+1}}$.

   Following the steps from above, we have shown that $\color{red}{(k+1)}\gt\color{red}{2^{k+1}}$.

4. Write a conclusion for your proof by induction.

   Answer (d):

   Hence, by mathematical induction we have proven that $n! \gt 2^{n}$ for $n ≥ 4$.