Question

Proof

Answer 1

Prove by Induction:

\(3^n > n!\)

\(\{n \in \mathbb Z : n \gt6 \}\)

Answer 2

Embarassing typo which explains, *maybe*, everything 😱

\(3^n \color{red}{\lt} n!\)

So we assume for n = k that:

\(P(k) := 3^k \lt k!\)

And we want \(P(k) \implies P(k+1)\)