Optimus prime number

If p^p-p=p, then p^p=2p, so p^(p-1)=2. We can see that p can be 2. Since 1 is not prime, the rest of the primes are integers greater than 2. So (2+k)^(2+k-1)=2, where k is an integer. So (2+k)^(k+1)=2. By binomial expansion (2+k)^(k+1)=2^(k+1)+(positive value). So this must equal 2 but for all k greater than 0 this will increase. Since all primes are integers greater than 2, they will all cause an increase to the computation. Since 2 is a number that satisfies this, all numbers greater than it will cause an increase and therefore not be equal to 2. Since 2 is the only integer satisfying this, 2 is also the only prime satisfying this which leads to our answer of 1 .

$\text{}{p}^{p}-p=p$

The above typicality just true for $\text{}({2}^{2})-2=2$ $\because$ only 2 has this typicality such that $\text{p}^{p}=p+p$

$\text{}(2^{2}=4)-2 =2$

$\large p^p - p = p \implies p^p = 2p.$

$p^p$ means multiplying $p$ by itself $p$ times. Now, we have $p^p = 2p$ . Where $p$ is multiplied by $2$ in the right hand side. So obviously, it must be $p = 2$ . Since 2 is a prime number, the answer is 1.