= Transcendental function
A <transcendental function> is an <analytic function> that is not an <algebraic function>. Explicitly, on a <domain> $U$, there is no nonzero <polynomial> $P(z,w)$ with complex coefficients for which
$$
P(z,f(z))=0\qquad\hbox{for every }z\in U.
$$
A <rational function> is algebraic, and a local branch of $\sqrt z$ is algebraic because it satisfies $w^2-z=0$. The <exponential function> is transcendental. This concerns a relation between functions, rather than the arithmetic nature of a single value: $e^0=1$ even though $z\mapsto e^z$ is transcendental.
Here is a direct proof for the <exponential function>. Suppose $\sum_{k=0}^m p_k(z)e^{kz}=0$ on $U$, with <polynomials> $p_k$ and $p_m\ne0$. The expression is an <entire function>, so the <identity theorem for holomorphic functions> extends the identity to all complex $z$. For positive real $x$, divide by $e^{mx}$:
$$
p_m(x)=-\sum_{k<m}p_k(x)e^{-(m-k)x}.
$$
By <exponential domination of polynomials>, the right side tends to zero as $x\to\infty$. A nonzero <polynomial> cannot tend to zero along the positive real axis, giving a contradiction. The case $m=0$ is already impossible because $p_0$ would vanish identically. Thus no such algebraic relation exists.
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