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.
Your proof is correct in its ideas, but contains a few unclear and even false statements. I'll list a few points of improvement:
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.
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.
This is again the same argument as before, which can be bundled with the previous arguments as well.
This is again the same argument as before, which can be bundled with the previous arguments as well.
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 $n=2^{2^k}$.$\quad\square$