Assume $a_k = k + 2$ for some arbitrary $k$, then show $a_{k+1} = (k+1) + 2$

Assume $a_k = k + 2$ for ALL $k \leq n$, then show $a_{n+1} = (n+1) + 2$ using the recurrence

Assume $a_n = n + 2$, then show $a_{n+1} = (n+1) + 2$ by direct computation

Use base cases $a_0 = 2$ and $a_1 = 3$ to verify the formula for all $n$