Roots of Unity

Is my solution correct?

Question:  Find all $z$, such that $​​​​(z+1)^7=z^7$.

My answer:  Since $z\ne 0$, we can divide by $z^7$ on both sides to get $\left(\frac{z+1}{z}\right)^7=1$. If we let $w=\frac{z+1}{z}$, then $w$ is equal to the $7^{\text{th}}$ roots of unity. We can manipulate $w=\frac{z+1}{z}$ by multiplying both sides by $z$, then subtracting $z$ from both sides to get $1=zw-z$. We can then divide both sides by $w-1$ to get $z=\frac{1}{w-1}$. 

The seven roots of unity are $1$, $e^{\frac{2i\pi}{7}}$, $e^{\frac{4i\pi}{7}}$, $e^{\frac{6i\pi}{7}}$, $e^{\frac{-6i\pi}{7}}$, $e^{\frac{-4i\pi}{7}}$, $e^{\frac{-2i\pi}{7}}$. $z=\frac{1}{1-1}$ if $w=1$ is not defined, and we can see that it doesn't work for $(z+1)^7=z^7$, either, so the solutions for $z$ are $\frac{1}{e^{\frac{2i\pi}{7}}-1}$, $\frac{1}{e^{\frac{4i\pi}{7}}-1}$, $\frac{1}{e^{\frac{6i\pi}{7}}-1}$, $\frac{1}{e^{\frac{-6i\pi}{7}}-1}$, $\frac{1}{e^{\frac{-4i\pi}{7}}-1}$, and $\frac{1}{e^{\frac{-2i\pi}{7}}-1}$, respectively.