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Show that if an entire function $f$ has a pole at $\infty$ of order $m$ then it is a polynomial of degree $m$.

My try:$f$ has a pole of order $m\implies g(z)=f(\frac{1}{z}) $ has a pole of order $m$ at $z=0\implies g(z)=z^m h(z)$ where $h(0)\neq 0$.

Also $f$ has a Taylor series Expansion around $0$ ,then $f(z)=a_0+a_1z+a_2z^2+\dots +a_nz^n+\dots$ ,

If I can show that $a_n=0\forall n> m$ then we are done.

Also $|f(z)|\to \infty $ as $|z|\to \infty \implies |f(z)|>M $ for $|z|>G$ .

Also $a_n=\dfrac{n!}{2\pi i}\int _R \dfrac{f(z)}{z^{n+1}}$ where $R$ is a positively oriented circle around $0$.

How can I show that $a_n=0$ for $n>m$ from here?

Please help.

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The Laurent series of $g(z) = f(\frac{1}{z})$ around 0 is given by $$ g(z) = a_0 + \frac{a_1}{z} + \frac{a_2}{z^2} + \dots,$$ where the $a_i$ are the coefficients from the power series expansion of $f$ around 0. Now use that $g$ has a pole of order $m$ at 0 and get $a_n = 0$ for $n > m$.

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