Characterization of primes of the form $n^n+1$ by using number-theoretic functions It is known that there is a unsolved problem related to primes of the form $n^n+1$ as is explained in page 160 of [1] (see also page 156, and the OEIS page related to this integer sequence A121270). In this post I ask about if we can to get a characterization for this sequence of primes in terms of (some) number-theoretic functions.

Question. I would like to get a characterization of these primes $n^n+1$ in terms of number-theoretic functions (see my attempts) It is required that your characterization is  a $\iff$ statement.

My attempt was the following deductions (while which I require is a $\iff$ statement: a characterization in terms of particular values of common arithmetic functions in number theory).
Claim 1. If $n^n+1$ is a prime number, then $$\varphi(\varphi(n^n+1))=\frac{\varphi(n)}{n}(\psi(n^n+1)-2)\tag{1}$$ holds, where $\varphi(k)$ denotes the Euler's Totient Function and $\psi(k)$ denotes the Dedekind Psi Function.
Claim 2. If $n^n+1$ is prime, then the equations
$$\varphi(\varphi(A))^B=(A-1)^{B-1}\cdot\varphi(B)^B$$
and $$\psi(A-1)^B=\psi(B)^B\cdot(A-1)^{B-1}$$
holds for some choice of integers $A,B\geq 1$ (take $B=n=\sqrt[B]{A-1}$).
Computation evidence for Claim 2. I've tested with a GP program (Sage Cell Server) that the only integers that satisfy both equations are $A=2,5$ and $257$ when the variables run over the integers $2\leq A\leq 10^3$ and $1\leq B\leq 10^3$
References:
[1] Michael Křižek, Florian Luca and Lawrence Somer, 17 Lectures on Fermat Numbers, CMS Books in Mathematics, Canadian Mathematical Society, Springer-Verlag (2001).
 A: Claims 1 and 2 are true. Let $n^n+1$ be a prime number. Then
$$
\begin{align}
\varphi(\varphi(n^n+1))&=\varphi(n^n)\\
&=n^{n-1}\varphi(n)\\
&=\frac{\varphi(n)}{n}n^n\\
&=\frac{\varphi(n)}{n}(n^n+2-2)\\
&=\frac{\varphi(n)}{n}(\psi(n^n+1)-2).
\end{align}
$$
which proves Claim 1. Let's call Claim 2a the statement
$$
\varphi(\varphi(n^n+1))^n=(n^n)^{n-1}\cdot\varphi(n)^n
$$
and call Claim 2b the statement:
$$
\psi(n^n)^n=\psi(n)^n\cdot(n^n)^{n-1}
$$
which are just the separate conjuncts of Claim 2 taking as you suggest $A=n^n+1$ and $B=n$. It's thus obvious that to prove Claim 2 it suffices to prove Claims 2a and 2b.
Proof of Claim 2a:
$$
\begin{align}
\varphi(\varphi(n^n+1))^n&=\varphi(n^n)^n\\
&=(n^{n-1}\varphi(n))^n\\
&=n^{n(n-1)}\varphi(n)^n\\
&=(n^n)^{n-1}\cdot\varphi(n)^n
\end{align}
$$
Proof of Claim 2b:
$$
\begin{align}
\psi(n^n)^n&=(n^{n-1}\psi(n))^n\\
&=\psi(n)^n\cdot(n^n)^{n-1}
\end{align}
$$
A: How good are you at power towers?
Given that $n^n+1$ must be a multiple of $n^m+1$ when $m$ is a divisor of $n$ and $n/m$ is odd, we can easily see that $n$ must be a power of $2$, thus $n=2^k$.
But wait, there's more. Suppose we try small values of $k$. We nicely get primes when $k=0,1,2$, but for $k=3$ we discover that $8^8+1$ is divisible by $257=2^8+1$, as well as $97$ and $673$.
If $n=2^k$ and $k=pq$ for some odd $p$, then
$n^n+1=2^{pq\cdot n}+1,$
and we conclude that our intended prime is in fact divisible by $2^{q\cdot n}+1$, which forces the number to be composite unless $p$ is forced to $1$ or $k=pq=0$. (The case we tried, $n=8$, corresponds to $p=3,q=1$.) So not only do we need $n=2^k$ for a prime, we also need $k=0$ or $k=2^l$ meaning the next candidate after $n=4$ is not until $n=16$. More broadly the only acceptable values of $n$ are given by the rapidly growing sequence
$1,2,4,16,256,...,$
where each term after those shown is the square of the previous term. The $n=4$ term is just $257$, but the $n=16$ candidate would already have more than twenty decimal digits.
We can express candidate function values greater than $1^1+1=2$ as Fermat numbers, specifically $\Phi_{2^l+l}$ for nonnegative whole numbers $l$. In terms of this the candidates defined above may be rendered:
$2,\Phi_1,\Phi_3,\Phi_6,\Phi_{11},\Phi_{20},\Phi_{37},\Phi_{70},...$
The first three of these entries are prime and the next four are known to be composite. So the next possible prime in this list is $\Phi_{70}$, corresponding to $n=2^{64}$. Although Pepin's Test offers a simplified means to assess primal-ity, the sheer size of this number prevents a computational solution in a reasonable time with current technology. A proof that all Fermat numbers above $\Phi_4$ are composite would solve the problem definitively.
