Let $F$ be a number field and let $\pi$ be a cuspidal automorphic representation of $GL_n(\mathbb{A_F})$.
Do we know that $L(s,\pi\times \tilde\pi)$ has a simple pole at $s=1$?
Do we know that $\frac{L(s,\pi\times \tilde\pi)}{\zeta_F(s)}$ is holomorphic at $s=1$?
Assuming that $\pi$ is cuspidal, then $L(s,\pi \times \tilde{\pi})$ always has a simple pole at $s = 1$. You can probably find this many places, but one such is this paper of Cogdell and Piatetski-Shapiro (see the Corollary on page 21, at the very end).
The answer to the second question is also yes, because the Dirichlet class number formula implies that $\zeta_{F}(s)$ also has a simple pole at $s = 1$.
Warning. I was thinking on $\pi$ not neccesarily cuspidal. Jeremy Rouse's answer is the correct one.
$L(s,\pi_1\times \pi_2)$ has a pole at $s=1$ if $\pi_2=\tilde\pi_1$, but it is not neccesarily simple.
I'm sure that $L(s,\pi\times \tilde\pi)/\zeta_F(s)$ is holomorphic for some $\pi$, but I suspect that the problem should be open in general. Note that we don't know if $\zeta_F(s)/\zeta(s)$ is an entire function for non-Galois $F$.
