Q.Prove by induction that $2^n > n^2$ for every integer $n \ge 5$.

[!TLDR] Base case shifts to $n=5$; double the hypothesis and show $2k^2\ge(k+1)^2$ for $k\ge5$. [!ANSWER] …

Let $P(n)$: $2^n>n^2$. Note $P(n)$ is false for $n=2,3,4$ (e.g. $2^4=16=4^2$, not $>$), so the base case is taken at $n_0=5$. Base case: $n=5$: $2^5=32$ and $5^2=25$, and $32>25$, so $P(5)$ holds. Inductive step: assume $P(k)$: $2^k>k^2$ for some $k\ge5$. Doubling: $2^{k+1}=2\cdot2^k>2k^2$. We need $2k^2\ge(k+1)^2=k^2+2k+1$, i.e. $k^2-2k-1\ge0$. For $k\ge5$: $k^2-2k-1=k(k-2)-1\ge5\times3-1=14>0$, so the inequality holds. Chaining: $2^{k+1}>2k^2\ge(k+1)^2$, so $2^{k+1}>(k+1)^2$, which is $P(k+1)$. By ind …

Inequality-by-induction with a shifted base case at $n_0=5$: double the hypothesis and verify the auxiliar …

Starting the base case at $n=1$ and getting confused when the statement fails at $n=3$ or $n=4$ -- always check the small cases first to fin …