Given $$ f(z)=\frac{z^{n}+1}{z}-z e^{\frac{1}{z}} $$ for $n \in \mathbb{N}$ and $z \in \mathbb{C}$
I am asked to find all the singulairiez points of $f(z)$
So from what I see, for every $n$ the only point is $z=0$ and since it is neither pole or removable singularity of $e^{\frac{1}{z}}$ it is neither pole or removable singularity of $f$
Is this argue true? or did I miss something here?