I have been doing some leisurely reading on complex analysis, and have came across several texts that mention the following:

* If an entire function has a removable singularity at infinity, then it must be constant
* If an entire function has a pole of order $k$ at infinity, then it must be a polynomial of degree at most $k$

I think this is a fascinating result - but I am struggling to figure out why exactly these are true.

I of course know that if $f$ has a removable singularity at infinity, then $g(z) = f(1/z)$ has a removable singularity at $0$. I have been playing around trying to get somewhere I can apply Liouville's Theorem, but can't quite get there.

For the second point, I know that if $f$ has a pole of order $k$ at infinity, then $g(z)=f(1/z)$ has a pole of order m at 0. This means there is some entire function $h$ where $g(z) = z^kh(z)$ and $h(0)\neq 0$. I was looking to bound this by a polynomial (as this would give the result), but not sure how to do this either...

Anyone able to help or provide proof would be hugely appreciated - keen to understand this better before reading further. :)

Thanks!