If $\alpha=(a_1a_2...a_k)$ is a $k$-cycle, then $\alpha^i(a_j)=a_{(j+i)\text{ mod }k}$. I am trying to show that if $\alpha=(a_1a_2...a_k)$ is a $k$-cycle, then for all $i\in\{1,2,...,k\}$ we have $\alpha^i(a_j)=a_{(j+i)\text{ mod }k}$. I was hoping folks could check my proof and perhaps comment on if it's more natural to use simple induction or strong induction.
Proof: (strong)
When $i=1$, if $j<k$ then by definition of the cycle $\alpha=(a_1a_2...a_k)$ we have $\alpha(a_j)=a_{(j+1)}=a_{(j+1)\text{ mod }k}$ since $j+1\leq k$ if $j<k$ and $a_{0 \text{ mod } k}= a_{k\text{ mod } k}$. If $j=k$, then again by definition of the cycle $\alpha=(a_1a_2...a_k)$ we have $\alpha(a_j)=\alpha(a_k)=a_1=a_{(k+1)\text{ mod }k}$.
Now assume by induction that for all $i$ less than $k$ and greater than $1$ that $\alpha^i(a_j)=a_{(j+i)\text{ mod }k}$ holds. Let $i=k$. Then $$\alpha^i(a_j)=\alpha^k(a_j)=\alpha(\alpha^{k-1}(a_j))$$ but by the induction hypothesis $\alpha^{k-1}(a_j)=a_{(j+k-1)\text{ mod }k}$ so $$\alpha^i(a_j)=\alpha^k(a_j)=\alpha(\alpha^{k-1}(a_j))=\alpha(a_{(j+k-1)\text{ mod }k}).$$

Now, since $(j+k-1)\text{ mod }k +1 =(j+k-1)\text{ mod }k+(1)\text{ mod }k=(j+k)\text{ mod }k$ and $(j+k-1)\text{ mod }k\leq k $ we have

again by definition of the cycle $\alpha=(a_1a_2...a_k)$ that:
$$\alpha(a_{(j+k-1)\text{ mod }k})=a_{(j+k)\text{ mod }k}=a_{(j+i)\text{ mod }k}$$
so $\alpha^i(a_j)=a_{(j+i)\text{ mod }k}$. Thus the claim holds for $i=k$, and by induction $\alpha^i(a_j)=a_{(j+i)\text{ mod }k}$ holds for any $k$-cycle where $k\in\mathbb{Z^+}$. $\blacksquare$
Proof: (simple)
When $i=1$, if $j<k$ then by definition of the cycle $\alpha=(a_1a_2...a_k)$ we have $\alpha(a_j)=a_{(j+1)}=a_{(j+1)\text{ mod }k}$ since $j+1\leq k$ if $j<k$ and $a_{0 \text{ mod } k}= a_{k\text{ mod } k}$. If $j=k$, then again by definition of the cycle $\alpha=(a_1a_2...a_k)$ we have $\alpha(a_j)=\alpha(a_k)=a_1=a_{(k+1)\text{ mod }k}$.
Now assume by induction that $\alpha^i(a_j)=a_{(j+i)\text{ mod }k}$ holds for some $i\in\{1,2,...,k\}$. Then by the induction hypothesis: $$\alpha^{i+1}(a_j)=\alpha(\alpha^i(a_j)=\alpha(a_{(j+i)\text{ mod }k})$$

so by definition of the cycle $\alpha=(a_1a_2...a_k)$ and that $(j+i)\text{ mod }k\leq k$ we have

$$\alpha(a_{(j+i)\text{ mod }k})=a_{(j+i+1)\text{ mod }k}.$$
Hence the claim holds for all $i$ in $\{1,2,...k\}$ and for all integers $k$. $\blacksquare$
Are there any gaps, specifically in the induction step(s) highlighted? I'm concerned I'm not properly accounting for the cases when the element is the last in the cycle $(=k)$. Thanks in advance!
 A: As for your main focus (expressed in a comment), let's start from the definition of $k$-cycle:
$\alpha(a_i)=a_{i+1}$, for $i=1,\dots,k-1$
$\alpha(a_k)=a_1$

Step to get insight on the formula to be proven by induction:
$\alpha^2(a_i)=\alpha(a_{i+1})=a_{i+2}$, for $i=1,\dots,k-2$
$\alpha^2(a_{k-1})=\alpha(\alpha(a_{k-1}))=\alpha(a_k)=a_1$
$\alpha^2(a_{k})=\alpha(\alpha(a_{k}))=\alpha(a_1)=a_2$
So, the proper definition of the $j$-th power of $\alpha$, $1\le j\le k$, relies on the splitting of the set $\{1,\dots, k\}$ into $\{1,\dots, k-j\}$ and $\{k-j+1,\dots, k\}$, which must be part of the induction process.

Induction:
Inductive hypothesis ("i.h."):

*

*$\alpha^j(a_i)=a_{i+j}$, for $i=1,\dots,k-j$

*$\alpha^j(a_{k-j+l})=a_l$, for $l=1,\dots,j$
For $j=1$ we retrieve the definition of $k$-cycle. Now the case $j+1$:

*

*for $i=1,\dots,k-(j+1)$, we get: $\alpha^{j+1}(a_i)=$ $\alpha(\alpha^j(a_i))\stackrel{(i.h.)}{=}$ $\alpha(a_{i+j})\stackrel{(j+1\le i+j\le k-1)}{=}a_{i+(j+1)}$;


*for $l=1,\dots,j+1$, we get: $\alpha^{j+1}(a_{k-(j+1)+l})=$ $\alpha(\alpha^j(a_{k-(j+1)+l}))=$ $\alpha(\alpha^j(a_{k-j+(l-1)}))$; now:

*

*for $l=1$, we get: $\alpha(\alpha^j(a_{k-j+(l-1)}))=$ $\alpha(\alpha^j(a_{k-j}))\stackrel{(i.h.,\text{ part 1})}{=}$ $\alpha(a_k)=a_1$;

*for $l=2,\dots,j+1$, we get: $\alpha(\alpha^j(a_{k-j+(l-1)}))\stackrel{(i.h.,\text{ part 2})}{=}$ $\alpha(a_{l-1})=$ $a_l$,

and hence $\alpha^{j+1}(a_{k-(j+1)+l})=a_l$ for every $l=1,\dots,j+1$. $\space\Box$

For $j<k$, from the part 1 follows $\alpha^j(a_1)=a_{j+1}\ne a_1$, and hence $\alpha^j\ne Id$.
On the other hand, for $j=k$, part 2 tells the whole story, which is: $\alpha^k(a_l)=a_l$, for $l=1,\dots,k$, namely $\alpha^k=Id$.
Therefore, $\left|\alpha\right|=k$.
