It makes sense to prove this by induction because after breaking the bar once, you are left with smaller chocolate bars. Reducing to smaller cases is what induction is all about. We can inductively assume we already know how to deal with these smaller bars. The problem is, if we are trying to prove the inductive case about a \((k+1)\)-square bar, we don't know that after the first break the remaining bar will have \(k\) squares. So we really need to assume that our conjecture is true for all cases less than \(k+1\text{.}\)
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