Proof.
Let \(P(n)\) be the statement \(2^n \lt n!\text{.}\) We will show \(P(n)\) is true for all \(n \ge 4\text{.}\) First, we check the base case and see that yes, \(2^4 \lt 4!\) (as \(16 \lt 24\)) so \(P(4)\) is true. Now for the inductive case. Assume \(P(k)\) is true for an arbitrary \(k \ge 4\text{.}\) That is, \(2^k \lt k!\text{.}\) Now consider \(P(k+1)\text{:}\) \(2^{k+1} \lt (k+1)!\text{.}\) To prove this, we start with the left side and work to the right side.
\begin{align*}
2^{k+1}~ \amp = 2\cdot 2^k \amp\\
\amp \lt 2\cdot k! \amp \text{by the inductive hypothesis}\\
\amp \lt (k+1) \cdot k! \amp \text{ since } k+1 \gt 2\\
\amp = (k+1)! \amp
\end{align*}
Therefore \(2^{k+1} \lt (k+1)!\) so we have established \(P(k+1)\text{.}\) Thus by the principle of mathematical induction \(P(n)\) is true for all \(n \ge 4\text{.}\)