A question about the asymptotes of tetration

[JmsNxn]

I hate to be a noob, but am I allowed to conclude that since \( sexp(-2) = ln(0) = -\infty \) that tetration has poles at all negative integers excluding -1?

I think I am, but some part of me is hesitant, I'm wondering if some people argue that the log law breaks down after zero. (Though I think that would be pretty stupid.)

You would be right in concluding there are singularities, but they're not poles -- they are \( \log^n \) singularities, i.e. the first is a logarithmic singularity, the second is a "double-logarithmic" singularity, and so on. In the complex numbers, \( \mathrm{tet}(z) \) is a "multi-valued function" (this term should really be something like multi-valued relation, but this misnomer is so ingrained in tradition it's not funny), like \( \log \) itself.

Okay, alright I'll make sure to not call them poles.