Proving $(x^s)^t=x^{st}$ 
Suppose $x\in G$ and $|x|=n<\infty.$ If $n=st$ for some positive integers $s$ and $t$, prove that $|x^s|=t.$

Observe that if we prove that $(x^s)^t=x^{st}$ then $1=x^n=x^{st}=(x^s)^t$ and we are done. We did not prove that $(x^s)^t=x^{st}$ in class and it is another exercise in the book. Is mathematical induction the only way to prove it?
My proof:
Statement: $(x^s)^t=x^{st}$ for all $1\leq s,t\leq n.$ 
Base Case: Trivially true.
Inductive Hypothesis: Assume that $(x^s)^t=x^{st}$ for all $1\leq s,t\leq n.$ 
Now consider (also assuming $x^{s+t}=x^sx^t$):
\begin{align}
(x^{s+1})^{t+1}\\
&=(x^{s+1})^t x^{s+1}\\
&=(x^sx)^tx^sx\\
&=(x^sxx^sx\cdots x^sx)x^sx\\
&=(x^s)^tx^tx^sx\\
&=x^{st}x^{s+t+1}\\
&=x^{st+s+t+1}\\
&=x^{(s+1)(t+1)}
\end{align} So by the principle of mathematical induction the statement is true.

If there are any mistakes please point them out.

 A: Others have addressed the main question, so I’ll just comment on your proof that $(x^s)^t=x^{st}$. You’re making it much too hard, and you’re not really taking advantage of induction.
Let $P(t)$ be the statement that $(x^s)^t=x^{st}$ for all $s\ge 1$. Clearly $(x^s)^1=x^s=x^{s\cdot1}$ for all $s$, so $P(1)$ is true. Now suppose that $P(t)$ is true for some $t\ge 1$. Then for any $s\ge 1$ we have
$$\begin{align*}
(x^s)^{t+1}&=(x^s)^tx^s\\
&\overset{(*)}=x^{st}x^s\\
&=x^{st+s}\\
&=x^{s(t+1)}\;,
\end{align*}$$
where the starred step uses the induction hypothesis $P(t)$. But $(x^s)^{t+1}=x^{s(t+1)}$ is precisely the statement $P(t+1)$, so we’ve shown that $P(t)$ implies $P(t+1)$, and by induction we can conclude that $P(t)$ is true for all $t\ge 1$. That is, for all $t\ge 1$ and for all $s\ge 1$, $(x^s)^t=x^{st}$, which was what you wanted to prove.
A: By definition $|x| = n$ means that $n$ is the smallest nonnegative integer such that $x^n = e$, where $e$ is the identity. 
If $|x^s| = u < t$, then $(x^s)^u = x^{su} = e$. Since $u < t$, $su < n$. This contradicts $n$ being smallest such that $x^n = e$. 
If $|x^s| = u > t$, then since $t < u$, one has $e \neq (x^s)^t = x^{st} = x^n$. So $x^n \neq e$. This contradicts $|x| = n$. 
Every element in a finite group has finite order, since we have shown that $|x^s| < t$ and $|x^s| > t$ is impossible, you must have $|x^s| = t$. 
A: I have two extended comments that won't fit in a comment on this site:
1) Are you doing a proof by induction on $s$ or on $t$. You need to fix one and use induction on the other. (Probably induction on $t$ will be easier.) When you do this, the calculations in the induction step will be easier.
2) With the minor fix suggestd in point 1), you have proved that $\left(x^s\right)^t=1$. However, this is not sufficient to conclude that $\left|x^s\right| = t$. You also need to prove that $t$ is the smallest positive integer such that $\left(x^s\right)^t=1$. This probably won't take much work, but it definitely needs to be included in your proof.
