Exercise 1.2.31 ($n-1$ divides $n^k - 1$)

Question

Let $n\geq 2$ and $k$ be be any positive integers. Prove that $(n-1) \mid (n^k - 1)$.

Richard Ganaye · Accepted Answer

\begin{proof} 
$$n^k - 1 = (n-1) \sum_{j = 0}^{k-1} n^j = (n-1) q,$$
where
$$q = \sum_{j = 0}^{k-1} n^j  = 1 + n + n^2 + \cdots +n^{k-1}$$
is an integer. So $n-1 \mid n^k - 1$.

(This remains true if $n = 0, 1$, or any integer in $\mathbb{Z}$.)
\end{proof}

Exercise 1.2.31 ($n-1$ divides $n^k - 1$)

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Exercise 1.2.31 ($n-1$ divides $n^k - 1$)

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