Here's what I have right now:
The order of a $k$-cycle in $S_n$ is $k$.
Proof. Let $\sigma$ represent the $k$-cycle $$\sigma=(x_1 \ x_2 \ \cdots \ x_k)$$with distinct elements $x_i$. Note that the cycle of distinct $x_i$ implies $k \leq n$. There are two cases to consider: $k$ even and $k$ odd. Suppose $k$ is even. It follows that $$\sigma^2=(x_1 \ x_2 \ \cdots \ x_k)(x_1 \ x_2 \ \cdots \ x_k)=(x_1 \ x_3 \ \cdots \ x_{k-1})(x_2 \ x_4 \ \cdots \ x_k)$$ $$\sigma^3=(x_1 \ x_2 \ \cdots \ x_k)\left[(x_1 \ x_3 \ \cdots \ x_{k-1})(x_2 \ x_4 \ \cdots \ x_k) \right]$$ $$=(x_1 \ x_4 \ \cdots \ x_{k-2})(x_2 \ x_5 \ \cdots \ x_{k-1})(x_3 \ x_6 \ \cdots \ x_k).$$ Thus, we can construct the general formula as a product of disjoint cycles $$\sigma^j=\prod_{i=1}^{j} (x_i \ x_{i+j} \ x_{i+2j}\ \cdots \ x_{k-j+i})$$ where $k$ is even. Now suppose $k$ is odd. Then $$\sigma^2=(x_1 \ x_2 \ \cdots \ x_k)(x_1 \ x_2 \ \cdots \ x_k)=(x_1 \ x_3 \ \cdots \ x_{k-2} \ x_k \ x_2 \ x_4 \ \cdots \ x_{k-1})$$ which differs from the case when $k$ is even. Interestingly, $$\sigma^3=(x_1 \ x_2 \ \cdots \ x_k)\left[(x_1 \ x_3 \ \cdots \ x_{k-2} \ x_k \ x_2 \ x_4 \ \cdots \ x_{k-1}) \right]$$ $$=(x_1 \ x_4 \ \cdots \ x_{k-2})(x_2 \ x_5 \ \cdots \ x_{k-1})(x_3 \ x_6 \ \cdots \ x_k)$$ which gives the same result as $k$ even. The reader can verify that all $\sigma^{j>2}$ are equal for $k$ even and $k$ odd. Thus, the general formula can be revised: $$\sigma^{j>2}=\prod_{i=1}^{j} (x_i \ x_{i+j} \ x_{i+2j} \ \cdots \ x_{k-j+i})$$ where $k>2$. For the case $k=2$, the $k$-cycle is a transposition which, by definition, has order 2. For the case $k=1$, the permutation is trivially the identity which has order 1. Thus, the theorem holds for $k=1,2$. For $k>2$, we know $$\sigma^{k}=\prod_{i=1}^{k} (x_i \ x_{i+k} \ x_{i+2k}\ \cdots \ x_{(k-k)+i}).$$ But for $n>0$ we see that $i+nk>k$ so it is not in the permutation. Thus, we simplify the above equation to $$\sigma^{k}=\prod_{i=1}^{k} (x_i)=e$$ We conclude that the order of a $k$-cycle in $S_n$ is $k$. QED.
The glaring problem seems to be my poor construction of the general case$$\sigma^{j>2}=\prod_{i=1}^{j} (x_i \ x_{i+j} \ x_{i+2j}\ \cdots \ x_{k-j+i})$$ which I considered replacing with$$\sigma^{j>2}=\prod_{i=1}^{j} (x_{i \bmod{j}})$$ but then I thought using this form is just as unenlightening as my original representation. I am just trying to convey that each disjoint cycle can only contain the elements $x_{i\leq k}$. A problem with using the $(x_{i \bmod{j}})$ notation is that we can run into a problem where elements of the form $x_{k<i\leq n}$ start appearing. Can you offer an elegant solution to this?
Also, I would appreciate advice on improving vague statements like this: "But for $n>0$ we see that $i+nk>k$ so it is not in the permutation." There are several lame statements like that in my proof that I just couldn't improve. I need help rewriting this proof in the most clear and concise manner.
Thank you!