Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)

Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.

Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and $y \geq 0$. Now suppose, for the sake of contradiction, $x \neq y$. Then $x > y$ or $x < y$. If $x > y$, then $0 \leq y < x$, so $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < y$ leads to a contradiction. Thus $x = y$. 
Case 2. $x \geq 0$ and $y < 0$. Now suppose, for the sake of contradiction, $x \neq -y$. Then $x > -y$ or $x < -y$. If $x > -y$, then $0 < -y < x$, so $(-y)^n < x^n$. Since $n$ is even, it follows that $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < -y$ leads to a contradiction. Thus $x = -y$. 
Case 3. $x < 0$ and $y \geq 0$. Applying case 2 with $x$ and $y$ interchanged, we get $y = -x$. Therefore $x = -y$. 
Case 4. $x <0$ and $y < 0$. Then $-x > 0$ and $-y > 0$. Applying case 1 to $-x$ and $-y$, we get $-x = -y$. Therefore $x = y$.
Is my proof correct? Is there a shorter way to prove this?
 A: The argument for Case 1 handles all four cases if you note that $x^n = |x|^n$ (etc.) when $n$ is even. (Just answerifying an earlier comment.)
A: The case $\ x=0,\ y=0\ $ is of the form $\ x=y$. 
Now without loss of generality set $\ y=c\cdot x$, which implies $\ c^n=1$, which implies $\ y=\pm\ x$.
A: If $x$ and $y$ are complex
(it is not stated that they are real),
this is false,
since $i^4 = 1^4 = 1$.
Just goes to show that
in math,
as in aerospace contracting,
it's always a requirements problem.
A: In general
$$x^n=y^n\iff 0=x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)$$
If $\,x\neq y \;$ then we must have that the second factor to the right above equals zero. 
But if $\,x,y >0\,$ ($\;x,y<0\;$), then that second factor is always positive (negative), so we can already focus on the case with different signs, and thus we can assume WLOG $\,x>0>y\;$ and the second factor's zero, thus:
$$x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}=P+N\;,\;\;\text{with}$$
$$P:=x^{n-1}+x^{n-3}y^2+\ldots+xy^{n-2}\;,\;\;N:=x^{n-2}y+x^{n-4}y^3+\ldots y^{n-1}$$
Observe that $\,P>0\;,\;N<0\;$ (why? Remember $\,n\;$ is even...) , so we can put
$$x^{n-1}+x^{n-3}y^2+\ldots+xy^{n-2}=-\left(x^{n-2}y+x^{n-4}y^3+\ldots y^{n-1}\right)\iff$$
$$x\color{red}{\left(x^{n-2}+x^{n-4}y^2+\ldots+y^{n-2}\right)}=-y\color{red}{\left(x^{n-2}+x^{n-4}y^2+\ldots y^{n-2}\right)}$$
End now the proof that it must be $\,x=-y\;$ ...
A: I would do it this way:
We have $x^n=y^n$ and $n=2k$ so we have $(y^2)^k=(x^2)^k$, taking $k$-th root of the both sides we obtain $y^2=x^2$ (beacuse $k$-th root function is an injection) so it follows $(x-y)(x+y)=0$ from which it follows that either $x=y$ or $x=-y$, short enough?
