A basic theorem of prime numbers. Checking My Proof 
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*Let $n$ be a positive integer. Prove that $n^5 -1$ is prime if and only if $n=2$.


$(\Rightarrow)$ Assume $n^5-1$ is prime. We will show $n=2$. If $n=3$, then clear $n^5-1$ is not prime.
Claim. If $n>2$, then $n^5-1$ is not prime.
Proof of the claim. We will do induction on $n$.
Initial Step. If $n=3$, clearly $n^5-1$ is not prime.
Induction Step. Assume for $n=m\in\mathbb{N}^{>2}$, $m^5-1$ is not prime. We will show $n=m+1\in\mathbb{N}^{>2}$, $(m+1)^5-1$ is not prime. So 
    $(m+1)^5-1=m((m+1)^4 +(m+1)^3 +(m+1)^2 +(m+1)+1)$
Let $u=m$ and $v=(m+1)^4 +(m+1)^3 +(m+1)^2 +(m+1)+1)$. Then since $m>2$, then $u>2$ and $v>2$. Thus, $uv$ cannot be a prime because $uv$ has no positive integer divisors other than $1$ and $uv$.
Therefore the claim is proved.
So by the claim, $n$ must be $2$.
$(\Leftarrow)$ Assume $n=2$. Then, $n^5 -1=2^5 -1=31$ is prime.
May you check my proof? Thanks...

Your proof is fine but really long and complicated and hard to read.
Simpler to just say:
$n^5 -1 = (n-1)(n^4 + n^3 + n^2 + n + 1)$.
If $n-1$ and $n^4 + n^3 + n^2 + n + 1$ are non trivial factors (not equal to $1$ or $n^5 -1$) then this is not prime.
So the only way for $n^5 -1$ to be prime is if either $n-1 = 1$ or $n^4 + n^3 + n^2 + n +1 = 1$.
If $n -1 = 1$ then $n =2$ and if $n^4 + n^3 + n^2 + n + 1= 1$ then $n(n^3 + n^2 + n + 1) = 0$ which has no natural number solutions.
So $n^5 -1$ is not prime if $n \ne 2$.
And if $n = 2$ then $n^5 -1 = 2^5 - 1 = 31$ which is prime.
