= Solution
A standard form of the <Schneider-Lang theorem> is this. Let $f_1,\ldots,f_s$ be meromorphic functions with $f_j'\in K[f_1,\ldots,f_s]$ for a fixed <number field> $K$. Suppose $f_1,f_2$ have <algebraic independence> over $\mathbb C$ and have finite growth orders $\rho_1,\rho_2$. There are at most $[K:\mathbb Q](\rho_1+\rho_2)$ points at which all the functions are finite and all their values lie in $K$.
Take $f_1(z)=z$ and $f_2(z)=e^z$, of orders zero and one. Their derivatives belong to $K[z,e^z]$. They have <algebraic independence>: a <polynomial> relation $\sum_jp_j(z)e^{jz}=0$ is impossible by exponential-polynomial independence, or by letting real $z$ tend to infinity and considering the largest $j$.
If nonzero algebraic $\alpha$ had algebraic $e^\alpha$, put $K=\mathbb Q(\alpha,e^\alpha)$. At each of the distinct points $z=n\alpha$, both $z$ and $e^z=(e^\alpha)^n$ belong to $K$. Infinitely many points contradict the theorem's finite bound. Therefore
$$
\boxed{e^\alpha\text{ is transcendental whenever }0\ne\alpha\in\overline{\mathbb Q}}.
$$
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