Verify my proof that for any $n>1$, if $n^n+1$ is prime, then $n=2^{2^k}$ for some integer $k$. I am solving a problem and I am respectfully asking someone to critique my work and offer suggestions on formatting or point out any glaring logical errors. Here is the problem:

Prove that for any $n>1$, if $n^n+1$ is prime, then $n=2^{2^k}$ for some integer $k$. Use this to prove that $2020^{2020}+1$ is not prime.

Proof. Let $n>1$ be an arbitrary positive integer such that $n^n+1$ is prime. We need to prove that $n$ must be of the form $2^{2^k}$ for some integer $k$. We see that if $n$ is odd, then $n^n$ is odd and thus $n^n+1$ is even, so $n$ must be even. Since $n$ is even, $n=2^pq$ for some integers $p$ and $q$, where $q$ is odd. We then see that $n^n+1=(2^pq)^{2^pq}+1=[(2^pq)^{2^p}]^q+1^q$. However, the only way that $n^n+1$ cannot be factored is if $q=1$, so $n=2^p$ for some integer $p$. If $p=1$, then we just notice that $p=2^0$, so $n=2^{2^0}$. Otherwise, if $p$ is odd and $p\neq 1$, then we see that $n^n+1=(2^p)^{2^p}+1=(2^{2^p})^p+1^p$, and since $p\neq 1$ and $p$ is odd, this expression can be factored, so $p$ must be even. This means that $p=2^kb$ for some integer $k$ and some odd integer $b$. We see that $n^n+1=(2^{2^kb})^{2^{2^kb}}+1=(2^{2^k2^{2^kb}})^b+1^b$, which can be factored if $b\neq 1$, so we see that $b=1$ and $p=2^k$. This means that $n=2^p=2^{2^k}$. Now, we see that $n^n+1=(2^{2^k})^{2^{2^k}}+1$, and we see that $2^k\cdot 2^{2^k}$ is even for all $k$, so $n^n+1$ cannot be factored and thus it is not necessary to go further. This means that if $n^n+1$ is prime, than $n=2^{2^k}$ for some integer $k$, completing the proof. $\blacksquare$
Now assume for the sake of contradiction that $2020^{2020}+1$ is prime. By our theorem, this means that $2020=2^{2^k}$ for some $k$. However, $2020$ is not a power of 2, contradiction. This means our assumption is false and $2020^{2020}+1$ is not prime.
 A: Your proof is correct in its ideas, but contains a few unclear and even false statements. I'll list a few points of improvement:

...if $n$ is odd, then $n^n$ is odd and thus $n^n+1$ is even, so $n$ must be even.

How does it follow that $n$ must be even? If you are concluding this because you have reached a contradiction from the assumption that $n$ is odd, I would recommend to state this explicitly.

We then see that $n^n+1=(2^pq)^{2^pq}+1=[(2^pq)^{2^p}]^q+1^q$. However, the only way that $n^n+1$ cannot be factored is if $q=1$...

Again I would recommend to be more explicit here; show that you can factor $n^n+1$ if $q>1$, and conclude that if $n^n+1$ is prime then $q=1$. In fact it is not true that $n^n+1$ cannot be factored if $q=1$; for $n=2^3$ you have
$$n^n+1=(2^3)^{2^3}+1=97\times257\times673.$$
Moreover, both arguments up to this point can be combined into a single argument. Instead of first proving that $n$ is even, simply write $n=2^pq$ with $p$ and $q$ nonnegative integers, where $q$ is odd.

...if $p$ is odd and $p\neq 1$, then we see that $n^n+1=(2^p)^{2^p}+1=(2^{2^p})^p+1^p$, and since $p\neq 1$ and $p$ is odd, this expression can be factored...

This is again the same argument as before, which can be bundled with the previous arguments as well.

We see that $n^n+1=(2^{2^kb})^{2^{2^kb}}+1=(2^{2^k2^{2^kb}})^b+1^b$, which can be factored if $b\neq 1$...

This is again the same argument as before, which can be bundled with the previous arguments as well.

Now, we see that $n^n+1=(2^{2^k})^{2^{2^k}}+1$, and we see that $2^k\cdot 2^{2^k}$ is even for all $k$, so $n^n+1$ cannot be factored and thus it is not necessary to go further.

It is not true that $n^n+1$ cannot be factored if $n=2^{2^k}$: For $k=4$ you have
$$n^n+1=(2^{2^4})^{2^{2^4}}+1=274177\times67280421310721.$$
In fact the sentence I quote above is redundant, no such thing asked in the question. Leaving it out would improve the proof. Here's a version of your proof with the suggested improvements:

Write $n=2^pq$ with $p$ and $q$ positive integers, and $q$ odd. Similarly write $p=2^kb$ with $k$ and $b$ positive integers, and $b$ odd, so that $n=2^{2^kb}q$ and
$$n^n+1=(2^{2^kb}q)^{2^{2^kb}q}+1.\tag{1}$$
Lemma. If $x$ and $y$ are positive integers and $y$ is odd, then $x^y+1$ is divisible by $x+1$.
Proof. Exercise.$\quad\square$
Taking $x=(2^{2^kb}q)^{2^{2^kb}}$ and $y=q$ shows that $n^n+1$ is divisible by $x+1>1$, so if $n^n+1$ is prime then $x+1=x^y+1$ and hence $y=q=1$. Then $(1)$ becomes
$$n^n+1=(2^{2^kb})^{2^{2^kb}}+1=2^{2^k2^{2^kb}b}+1.\tag{2}$$
Taking $x=2^{2^k2^{2^kb}}$ and $y=b$ shows that $n^n+1$ is divisible by $x+1>1$, so if $n^n+1$ is prime then $x+1=x^y+1$ and hence $y=b=1$, and so $n=2^{2^k}$.$\quad\square$
