in Theory of Computation
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Use induction on $n$ to show that $|u^n|=n|u|$ for all strings $u$ and all $n$.
in Theory of Computation
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Solution from Peter Linz itself:

For $n = 1,$ $|u^1| = |u|.$ Assume $|u^k| = k|u|$ holds for $k = 1, 2, ..., n,$ then

$|u^{n+1}| = |u^nu| = |u^n| + |u| = n|u| + |u| = (n + 1)|u|.$

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