This is exercise 2-9 from Spivak's Calculus:
Prove that if a set $A$ of natural numbers contains $n_0$ and contains $k + 1$ whenever it contains $k$, then $A$ contains all natural numbers $> n_0$.
Spivak's solution goes something like this: Let $B = \{l \in \mathbb{N} \mid n_0 + l \in A \}$. Clearly $0 \in B$. Now suppose $l \in B$. This means $n_0 + l \in A$. By the property of $A$ it follows that $n_0 + l + 1 \in A$ and thus $l+1 \in B$. Therefore, $B = \mathbb{N}$ and $A$ contains all $n \ge n_0$.
My attempted solution, however, was more mechanical. I will readily admit that defining $B$ to be an inductive set in terms of $A$ is way more elegant, but I still wanted to run my proof by math.se to see if it has any flaws in its logical construction. This is my proof:
Let $P(n) := n\ge n_0 \to n\in A$. For the inductive hypothesis, suppose $P(n)$ is true, and suppose $n+1 \ge n_0$. If $n+1 = n_0$ then clearly $n+1 \in A$. If $n+1 > n_0$, then $n \ge n_0$, so by assumption $n \in A$. By the property of $A$ it also follows that $n+1\in A$. Hence $P(n+1)$ is also true. For the base case, if $n_0 > 0$, then $P(0)$ is vacuously true. If $n_0 = 0$, then $0 \in A$ and $P(0)$ is true as well. Therefore, $\forall n \ge n_0(n\in A)$.
The interesting question is if this base case proof works correctly.