What do you think about my first proof employing strong induction? What mark/grade would you give me?
Theorem
Every natural number greater than 1 is a product of one or more primes.
Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephrase the theorem as follows. $$\forall n \in \{2, 3, \dots \}, \, P(n) \quad \text{where} \quad P(n) \, := \, n \text{ is a product of one or more primes}$$ We prove the theorem by strong induction; we induct on $n$.
Basis
We have $P(n)$ for $n = 2$, since $2$ is a product of one prime, namely $2$.
Inductive step
Below, we show that for all $n \in \{2, 3, \dots \}$, $$P(2) \land \dots \land P(n - 1) \land P(n) \Rightarrow P(n + 1)\text{.}$$
Let $k \in \{2, 3, \dots \}$. We assume that $P(2) \land \dots \land P(k - 1) \land P(k)$ is true. In the following, we use this assumption to show that $P(k + 1)$ is true.
We consider $k + 1$. There are two cases: It is prime or not. Obviously, the cases are exhaustive. We consider the cases separately; in each consideration, we conclude that $P(k + 1)$ is true. In so doing, we conclude the inductive step for all cases.
Case 1: Obviously, $k + 1$ is a product of one prime, namely $k + 1$. Thus $P(k + 1)$ is true.
Case 2: As $k + 1$ is not prime, by definition, it may be represented as a product of at least two natural numbers that are greater than $1$. Thus there are $i, j \in \{2, \dots, k - 1, k\}$ such that $ij = k + 1$. By the induction hypothesis, $P(i)$ and $P(j)$ are true. Hence, we can easily see that the product $\,ij\,$ is a product of primes. Thus $P(ij)$ is true. (Thus $P(k + 1)$ is true.)
By the basis, the inductive step, and the method of strong induction; $P(n)$ is true for all $n \in \{2, 3, \dots \}$.