Q.Prove by induction that $n! \ge 2^{n-1}$ for every natural number $n$.

[!TLDR] Multiply the inductive hypothesis by $(k+1)\ge2$. [!ANSWER] Proved: $n!\ge2^{n-1}$ for every natural number $n$.

Let $P(n)$: $n!\ge2^{n-1}$. Base case: $n=1$: $1!=1$ and $2^{0}=1$, so $1\ge1$ and $P(1)$ holds. Inductive step: assume $P(k)$: $k!\ge2^{k-1}$ for some $k\ge1$. Then $(k+1)!=(k+1)\cdot k!\ge(k+1)\cdot2^{k-1}$. Since $k\ge1$, $(k+1)\ge2$, so $(k+1)\cdot2^{k-1}\ge2\cdot2^{k-1}=2^k$. Chaining: $(k+1)!\ge(k+1)2^{k-1}\ge2^k$, so $(k+1)!\ge2^k$, which is $P(k+1)$. By induction, $P(n)$ holds for all $n\ge1$. [!ANSWER] $n!\ge2^{n-1}$ for every natural number $n$.

Inequality-by-induction: multiply the inductive hypothesis by $(k+1)$ and use $(k+1)\ge2$ for $k\ge1$.

Using the strict inequality $n!>2^{n-1}$ instead of $\ge$ -- the two sides are actually equal at $n=1$ and $n=2$, so only the non-strict version is true for all $n\ge1$.