Why is the emboldened statement in this question about products of disjoint cycles correct?

Question

Question:

Let $\alpha$ be an arbitrary permutation in the symmetric group $S_{n}$.

One of the cycles which composes $\alpha$ can be represented as the sequence $1, 1\alpha, 1\alpha^{2},1\alpha^{3}, 1\alpha^{4},\cdots$.

Suppose $1\alpha^{m}$ is the first term which is a repetition of a preceding term in the sequence, and that in fact $1\alpha^{m}=1\alpha^{k}$ with $m>k\geq0$. Then $1\alpha^{m-1}=1\alpha^{k-1}$ and this contradicts $1\alpha^{m}$ being the first repetition, unless $k=0$.

Deduce that the sequence has exactly $m$ different digits in it and that these $m$ digits are repeated in the same order.

These $m$ digits form one of the cycles of $\alpha$.

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I'm assuming $1\alpha^{m-1}=1\alpha^{m}\cdot\alpha^{-1}=1\alpha^{k}\cdot\alpha^{-1}=1\alpha^{k-1}$ is how the contradiction was arrived at. But can't the same argument be used to show $1\alpha^{m-2}=1\alpha^{k-2}$?

Answer 1

Of course the same argument can be used to show $1\alpha^{m-2}=1\alpha^{k-2}$, if $k\ge 2$. The point is to show that $k=0$, if not then $k>0$, so $1\alpha^{k-1}$ appears in the sequence and $1\alpha^{m-1}=1\alpha^{k-1}$ hence the contradiction.