On Proving The nth odd number is 2n − 1 Through Induction, And A Few Extensions

Question

What I've done so far is to prove it for the n+1th number:

$(2n-1)+2=2(n+1)-1$. Because any odd number +2 is equals to the next odd number. And in the proof, it is given that 2n-1 is an odd number.

$2n+1=2n+2-1$

$2n+1=2n+1$

What I'm worried though, is if by assuming "since any odd number +2 is equals to the next odd number", am I being circular? Am I using circular reasoning?

And also, thinking about this, how would one prove that for any odd number n, the next odd number is n+2? How would they do that without being circular?

Please give the answer very simple, at the level of a pre-calculus student. I am just starting proofs for fun, and am still at a very basic level. I haven't even touched on sigma notation yet.

Answer 1

$2n – 1 = 2n -2 + 1 = 2(n – 1) + 1 = 2k + 1$ where $k = n-1$ is an integer and $ 2n -1$ is odd by definition

Answer 2

$2n\ -\ 1$ is clearly odd because $(2n\ -\ 1)\ \text{mod}\ 2\ =\ 1$. So $2n\ -\ 1$ is an arbitrary odd number by definition. But adding 2 results in $2n\ +\ 1$ which is odd for the same reason. Hence, adding 2 to an arbitrary odd number results in another odd number, specifically the next consecutive odd number.