Order of a group element. [$\operatorname{ord}(x)$]

$\newcommand{\ord}{\operatorname{ord}}$

I need to prove the following

$(i)$ $\forall x\in G\,:\ord(x)=\ord(x^{-1})$
$(ii)$ $\forall x,g\in G\,:\ord(gxg^{-1})=\ord(x)\qquad$ (for $\ord(x)<\infty$ and $\ord(x)=\infty$)

proof $(i)$: let $\ord(x)=n$ and let $\ord(x^{-1})=m$. \begin{align} \ord(x)=n&\Leftrightarrow x^n=e\tag{1}\\ &\Rightarrow x^n \cdot (x^{-1})^n=(x^{-1})^n\\ &\Rightarrow e=(x^{-1})^n \end{align} Therefore $m=\ord(x^{-1})\le \ord(x)=n$
And \begin{align} \ord(x^{-1})=m&\Leftrightarrow (x^{-1})^m=e\tag{2}\\ &\Rightarrow (x^{-1})^m \cdot ((x^{-1})^m)^{-1}=x^{m}\\ &\Rightarrow e=x^{m} \end{align} Therefore $m=\ord(x^{-1})\ge \ord(x)=n$. So we know that $\ord(x^{-1})=\ord(x)$

proof $(iia)$: Let $\ord(x)=n$ and let $\ord(gxg^{-1})=m$ and assume that $\ord(x)<\infty$ \begin{align} \ord(gxg^{-1})=m\Leftrightarrow(gxg^{-1})^m&=e\tag{3}\\ \end{align} This is the same as \begin{align} \prod_{k=1}^{m}{(gxg^{-1})}&=gxg^{-1}\cdots gxg^{-1}\tag{4}\\ &=gx^mg^{-1}\\ &=e \end{align} So we know that \begin{align} &gx^mg^{-1}=e\\ &\Rightarrow gx^m=g\\ &\Rightarrow x^m=g^{-1}g\\ &\Rightarrow x^m=e \end{align} Therefore $n=\ord(x)\le \ord(gxg^{-1})=m$
We can do the same for \begin{align} &x^n=e\tag{5}\\ &\Rightarrow gx^m=g\\ &\Rightarrow gx^mg^{-1}=e \end{align} Therefore $n=\ord(x)\ge \ord(gxg^{-1})=m\Longrightarrow n=m$

proof $(iib)$: And here I get stuck.

My question:

How can i prove $\ord(gxg^{-1})=\ord(x)\qquad$ if $\ord(x)=\infty$

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Suppose $\operatorname{ord}(x)= \infty$.

Then $\forall n \in \mathbb N: x^n \neq e$, and $gxg^{-1}= gx^ng^{-1}$. Suppose $\operatorname{ord}(gxg^{-1}) = n$. Then $gx^ng^{-1} = e$, therefore $x^n = e$. So $\operatorname{ord}(gxg^{-1})$ has to be $\infty$.

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Hint: Show that $(gxg^{-1})^m = gx^mg^{-1}$ (you can use induction). Now assume for contradiction $gxg^{-1}$ has finite order. Then $gx^mg^{-1} = e$ for some $m \in \mathbb{N}$. Then what is $x^m$?

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Suppose $x$ has infinite order, and for contradiction suppose $gxg^{-1}$ has finite order $n$. Then $(gxg^{-1})^n = gx^ng^{-1} = 1$ hence $x^n = 1$, a contradiction.

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