= Solution
Use the local coordinate $w=1/z$ at infinity. For positive integer $N$,
$$
f(1/w)=w^{-N}e^{-w}=\sum_{m=0}^\infty\frac{(-1)^m}{m!}w^{m-N}.
$$
The negative powers of this local coordinate are precisely $m=0,\ldots,N-1$. The singular part at infinity is therefore
$$
\boxed{\sum_{m=0}^{N-1}\frac{(-1)^m}{m!}w^{m-N}=\sum_{m=0}^{N-1}\frac{(-1)^m}{m!}z^{N-m}.}
$$
Infinity is a <pole at infinity> of order $N$. In the original $z$ variable these singular terms are positive powers; it is the negative powers of $w=1/z$ that define the local <principal part of a Laurent series>. For $N=0$ the point at infinity is regular, and for negative integer $N$ it is a zero rather than a pole.
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