Invert $\sin(z)$ using geometric series

Just a computation: \[ {1\over \sin(z)} &= {1\over z-{z^3\over 3!} + {z^5\over 5!} - \bigo(z^7) } \\ &={1\over z\qty{ 1 - {z^2\over 3!} + {z^4 \over 5!} - \bigo(z^6)} } \\ &= \inverseof{z} \qty{1\over 1 - p(z)} \qquad p(z) \da {z^2\over 3!} - {z^4\over 5!} + \bigo(z^6) \\ &= \inverseof{z} \sum_{k\geq 0} p(z)^k \\ &= \inverseof{z}\qty{ 1 + p(z) + p(z)^2 + \bigo(z^2)^3 } \\ &= \inverseof{z}\qty{ 1 \ + \qty{{z^2\over 3!} - {z^4\over 5!} + \bigo(z^6)} \ + \qty{{z^2\over 3!} - {z^4\over 5!} + \bigo(z^6)}^2 \ + \bigo(z^6)} \\ &= {1\over z}\qty{ 1 + {1\over 3!}z^2 + \qty{\qty{1\over 3!}^2 - {1\over 5!} }z^4 + \bigo(z^6) }\\ &= {1\over z} + {1\over 6}z + {7\over 360}z^3 + \bigo(z^5) .\]