What about \(n=4\text{?}\) Now we could have a \(2\times 2\) bar, or a \(1 \times 4\) bar. In the first case, break the bar into two \(2\times 2\) bars, each which require one more break (that's a total of three breaks required). If we started with a \(1 \times 4\) bar, we have choices for our first break. We could break the bar in half, creating two \(1\times 2\) bars, or we could break off a single square, leaving a \(1\times 3\) bar. But either way, we still need two more breaks, giving a total of three.
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