= Solution
The <Cauchy-Riemann equations> are $u_x=v_y$ and $u_y=-v_x$. A <holomorphic function> is smooth, so equality of mixed <partial derivatives> gives
$$
\boxed{\Delta u=(v_y)_x-(v_x)_y=0,\qquad\Delta v=-(u_y)_x+(u_x)_y=0.}
$$
Thus its real and imaginary parts are <harmonic functions>.
An <entire function> is holomorphic at every finite point of the complex plane. The globally convergent <power series> for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore gives
$$
\sinh z=\frac{e^z-e^{-z}}2,\qquad (\sinh z)'=\cosh z
$$
at every complex $z$, proving that the <hyperbolic sine> is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $nz^{n-1}$, the <product rule> gives $(z^{n+1})'=z^n+z\,nz^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.
A function has a <pole> of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0)^nf(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne0$. An isolated <essential singularity> is neither a removable singularity nor a pole; equivalently its <Laurent series> has infinitely many nonzero negative-power terms. In particular
$$
e^{1/z}=\sum_{m=0}^\infty\frac{z^{-m}}{m!},\qquad 0<|z|<\infty,
$$
has an essential singularity at zero.
For $f_N$, the <Laurent series> about infinity is
$$
\boxed{f_N(z)=\sum_{m=0}^\infty\frac{z^{N-m}}{m!}=z^N+z^{N-1}+\frac{z^{N-2}}{2!}+\cdots+\frac1{N!}+\frac{z^{-1}}{(N+1)!}+\cdots.}
$$
To classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}e^w$, and $e^w$ is holomorphic with value one at zero. Therefore
$$
\boxed{f_N\text{ has a pole of order }N\text{ at infinity}.}
$$
The <pole at infinity> is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.
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