<s>[INST] Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. Full Problem: Prove that for any element $x$ in a group $G$ that satisfies 
$$x^9 = e \\
x^{11} = e,$$
where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that


*

*$x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$

*$x^{9} = x^{2} \cdot x^{7} = e \cdot x^7 \Rightarrow x^7 = e$

*$x^{7} = x^{2} \cdot x^{5} = e \cdot x^5 \Rightarrow x^5 = e$

*$x^{5} = x^{2} \cdot x^{3} = e \cdot x^3 \Rightarrow x^3 = e$

*$x^{3} = x^{2} \cdot x = e \cdot x \Rightarrow x = e$


Therefore, $x = e$.
 [/INST] $$e=(x^{11})^5=x^{55}=x^{54}\cdot x=(x^9)^{6}\cdot x=e\cdot x=x$$
As you can see, this thus follows because there is integer solutions to $11x-9y=1$, which is true because $11$ and $9$ are relatively prime.
Your approach is doing much the same, using a slow form of the Euclidean algorithm to show that $11$ and $9$ are relatively prime:
$$11=9\cdot 1 + 2\\
9=2\cdot 1 + 7\\
7=2\cdot 1 + 5\\
5=2\cdot 1 + 3\\
3=2\cdot 1 + 1$$
You could have skipped a lot of steps by doing the equivalent of $9=2\cdot 4 + 1$, as other answers have suggested.
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