Is induction better than other proof methods? So, I got the following question asked on a homework:

Let $a,b\in\mathbb{Z}$ and $m,n\in\mathbb{N}$. Show that $(a,b)=1\iff (a^n,b)=1$, and also $(a,b)=1\iff (a^n,b^m)=1$.

And here's my try...
Proof of $(a,b)=1\iff(a^n,b)=1$.
$\implies)$ Suppose $(a,b)=1$. Therefore, there exist $s,t\in\mathbb{Z}(as+bt=1)$. Note that $1^n=(as+bt)^n=1$. $$\therefore\sum_{k=0}^{n}{n\choose k}(as)^{n-k}(bt)^k=1.\\ \therefore a^n(s^n)+b\left(\sum_{k=1}^{n}{n\choose k}(as)^{n-k}{\space}b^{k-1}t^k\right)=1.$$
Since $1$ is a linear combination of $a^n$ and $b$, then $(a^n,b)=1$.
$\impliedby)$ Suppose $(a^n,b)=1$. Therefore, there exist $s,t\in\mathbb{Z}(a^ns+bt=1)$. Then, $a(a^{n-1}s)+bt=1$.
Since $1$ is a linear combination of $a$ and $b$, then $(a,b)=1$.
Proof of $(a,b)=1\iff(a^n,b^m)=1$.
From the proposition shown above, it follows that $(a,b)=1\iff (a^n,b)=1$.
Then, $(a^n,b)=(b,a^n)=1\iff (b^m,a^n)=1=(a^n,b^m)$, making use of the proposition once more.

Now, I have two questions.
First: is this proof correct? for some reason, I feel like I must use induction to prove this formally (the first proposition). Also, I feel like the second proof only holds when proving the first proposition by induction, but I don't know why (it's just a feeling).
Second: is induction better? If so, why? I feel like the first proof should be fine, but at the same time, I feel like I'm doing something wrong... that I'm not proving it for all natural numbers for not using induction. But, then again, we use this strategy of "choosing an arbitrary number in a set to prove it's true for all numbers in that set," so I don't see why it shouldn't hold.
I don't know if I'm making myself clear enough with my questions, but I'll leave it like this for now.
 A: Your proofs look perfectly good to me, and there's nothing about the second part that requires the first one to have been proved by induction.
Induction is not in and of itself a "better" way to prove things. There are two different factors that (each in their way) can make it feel like that at times:

*

*You have probably had homework exercises where you were explicitly asked to prove such-and-such by induction, and therefore proving it in another way was a wrong solution to that homework. However, that's because the point of those exercises was not really to get such-and-such proved, but to give you an opportunity to practice induction. That particular homework was more about the journey than the destination -- but nobody says you need to stay in that mode forever.


*In a fundamental sense, proving anything interesting about all the natural numbers needs an induction argument to be present somewhere. That's at least true if you whittle everything down to a fully formal argument resting on a mainstream foundation for mathematics -- because in those mainstream foundations, the only thing we really know about the natural numbers by definition is that they're the set that mathematical induction works for!
However, this doesn't mean than this underlying use of induction has to be visible on the surface of your proof, or even that it's in some sense "better" for it to be. A proof that builds on an intermediate result that you proved previously using some sort of induction -- such as the binomial theorem -- is fully as good as one where you do everything from scratch. In fact, arguably a proof that uses known concepts (in a well-motivated way) is strictly better than one that does everything from scratch just because.
A: From what I see of your answer (I haven't tried that question before, but from a first-sight perspective) I guess your first proof is correct (frankly, I didn't check the second). Even I am a beginner, so I'll be ready to review the proof a bit later and tell you if it's actually correct (I have a class to attend and so I don't prefer to take myself into checking this now).
Now, to your second question : you needn't think like that. As the kind of question changes, you'll have to think up different ways. For example, to prove that the perpendicular from a point onto a line is the shortest distance between them requires you to use the way of contradiction. Why induction works is because if the statement given to prove is truly applicable to the domain it acts on, then you will one way or the other get to the right conclusion that the statement holds in the given domain. So, overall, the method of proof depends in the kind of question given. For example, in my view, the simplest way of proving that $\nexists p \in \mathbb{P} : p \mid \sqrt{n }, p \leq \sqrt{n}, n\in \mathbb{N} \implies n \in \mathbb{P}$ and that $(a,b) = 1 \implies (a^n,b) = 1$ are entirely different. Probab;y the former can be proved using induction, but as far as I see, the type of proof is dependent on the  question.
