= Simply periodic entire function without an algebraic addition theorem
{title2=$f(z)=e^{e^z}$}
The displayed <entire function> has exactly the periods $2\pi i\mathbb Z$: a period $c$ would make $e^z(e^c-1)$ a constant integer multiple of $2\pi i$, forcing $e^c=1$. If a <polynomial> addition relation existed, specialize its middle argument at a point with $e^c=\alpha>0$ irrational, avoiding the finitely many specializations making the entire <polynomial> vanish. With $t=e^z$, the relation becomes a nonzero finite sum of exponentials $e^{(m+\alpha n)t}$. These real exponents are distinct, so the largest one dominates as $t\to+\infty$, a contradiction. This demonstrates that an <algebraic addition theorem> is stronger than simple periodicity.
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