Need help understanding a proof about permutations in Rotman's ''Advanced Modern Algebra'' I need a hand in understanding the proof from "Advanced Modern Algebra" by Joseph J. Rotman of the following theorem:

Let $\alpha \in S_n$ and let $\alpha = \beta_1...\beta_t$ be a complete factorization into disjoint cycles. This factorization in unique except for the order in which the cycles occur.

First, I would like to provide definitions and facts assumed here. There aren't a lot of them.

1)Disjoint permutations $\alpha, \beta \in S_n$ commute.
2)Every permutation $\alpha \in S_n$ is either a cycle of a product of disjoint cycles
3)A complete factorization of a permutation $\alpha$ is a factorization of $\alpha$ into disjoint cycles that contains exactly one $1$-cycle $(i)$ for every $i$ fixed by $\alpha$.
4)There is a relation between the notation for an $r$-cycle $\beta = (i_1 i_2...i_r)$ and its powers $\beta^k$, where $\beta^k$ denotes the composite of $\beta$ with itself $k$ times. Note that $i_2=\beta(i_1), i_3 = \beta^2(i_1), i_4=\beta^3(i_1)$, and
$\forall k, 0<k<r \ \ i_{k+1}=\beta^k(i_1)$
5)The inverse of the cycle $\alpha = (i_1i_2...i_r)$ is the cycle $(i_ri_r-1...i_1)$
6)If $\gamma \in S_n$ and $\gamma = \beta_1...\beta_k$, then $\gamma^{-1}=\beta^{-1}_k...\beta^{-1}_1$

Now, the proof:

Since every complete factorization of $\alpha$ has exactly one $1$-cycle for each $i$ fixed by $\alpha$, it suffices to consider( not complete) factorizations into disjoint cycles of lengths $\geq 2$. Let $\alpha = \gamma_1...\gamma_s$ be a second factorization of $\alpha$ into disjoint cycles of lengths $\geq 2$
The theorem is proved by induction on $l$, the larger of $t$ and $s$. The inductive step begins by noting that if $\beta_t$ moves $i_1$, then $\beta^k_t(i_1) = \alpha^k(i_1)$ for all k $\geq 1$. Some $\gamma_j$ must also move $i_1$ and, since disjoint cycles commute, we may assume that $\gamma_s$ moves $i_1$. It follows that $\beta_t=\gamma_s$; right multiplying by $\beta^{-1}_t$ gives $\beta_1...\beta_{s-1}$, and the inductive hypothesis applies.

So, what do I ask from you? Well, you could try to reformulate the proof so it will be more clearer. Or provide your own proof, perhaps.
At the very least, I need to understand how exactly an induction is applied here. What is an inductive step? I don't see it in the proof unfortunately.
Thank you in advance.
 A: Let $\alpha$ be any permutation. Suppose we have two complete factorizations
$$\alpha=\beta_1\dots\beta_t,$$
$$\alpha=\gamma_1\dots \gamma_s.$$
Let $l=\operatorname{max}(s,t)$. We induct on $l$. Explicitly, suppose we have shown the following: given any permutation and two factorizations of that permutation (as above) such that the maximum of their lengths is $n$ or less, the factorizations must be the same (up to order). We will prove the statement for $n+1$. (You should also think about the base case, but I omit that here.)
We may suppose $\alpha$ is not the identity (or there is nothing to prove). By relabeling, we may suppose $\beta_t$ is at least length two. (If all cycles have length one, then $\alpha$ is the identity and the statement is trivial.) So $\beta_t$ has length at least two and therefore moves some element, say $i_1$. Since the cycles in the factorizations are disjoint, the only part of the $\beta$ factorization that moves $i_1$ is $\beta_t$. So we have
$$\alpha^k(i_1)=\beta_t^k(i_1)$$
for all $k\ge 1$. Since $\alpha$ moves $i_1$, some cycle in the $\gamma$ factorization must move $i_1$. By relabeling the $\gamma_i$, we may assume it is $\gamma_s$. So 
$$\gamma_s^k(i_1)=\beta_t^k(i_1)$$
for all $k\ge 1$.
Since $\gamma_s$ and $\beta_t$ are both cycles, they are determined by their action on the single element $i_1$. (Think about why this is true.) So we conclude that $\gamma_s=\beta_t$. So
$$\alpha\gamma_s^{-1}=\beta_1\dots\beta_{t-1}=\gamma_1\dots \gamma_{s-1}.$$
Now we can apply the induction hypothesis to these smaller factorizations, which implies they are the same (up to order). Then we see, multiplying both sides by $\gamma_s=\beta_t$, that both factorizations of $\alpha$ are the same, and we're done.
