Q.Prove that $2^n > n$ for all positive integers $n$.

[!TLDR] Double the inductive hypothesis $2^k>k$ and use $2k\ge k+1$ for $k\ge1$. [!ANSWER] Prove …

Let $P(n)$: $2^n>n$. Base case: $n=1$ gives $2^1=2>1$, so $P(1)$ holds. Inductive step: assume $P(k)$: $2^k>k$ for some $k\ge1$. Multiplying both sides by $2$: $2^{k+1}=2\cdot2^k>2k$. Since $k\ge1$, we have $2k=k+k\ge k+1$. Chaining: $2^{k+1}>2k\ge k+1$, so $2^{k+1}>k+1$, …

Inequality-by-induction: double the inductive hypothesis, then chain with …

Stopping at $2^{k+1}>2k$ without the extra auxiliary step $2k\ge k+1$ needed to reach the target $k+1$ -- the c …