Why isn't mathematical induction a circular argument? In the induction step, we need to show the thing we're trying to prove is true when n=k+1, using our assumption that it's true when n=k. But how can we truly prove a thing by using an assumption?
 A: The crucial thing that you've missed about the induction, is that it has to be true for $n=0$ case. Without that, you are correct, you only chain assumptions with conclusions, which does not mean that the whole statement is true.
Let me give you a concrete example and counterexample:
Example. Say we want to prove that $0+1+2+\cdots +n=n\cdot (n+1)/2$ for any natural $n$. There are two steps involved in the induction:

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*Case $n=0$. Then we have $0$ on the left side and $0\cdot (0+1)/2=0$ on the right side. These two are equal and so the statement is simply true for $n=0$.

*Now assume that the statement is true for some $n$. We will show that it is true for $n+1$. Indeed, for $n+1$ we have $0+1+2+\cdots +n+(n+1)$ on the left side. This is equal to $(0+1+2+\cdots +n)+(n+1)$, I've only inserted brackets here. But we know what this smaller sum is, because we assume that the statement is true for $n$. Thus the expression is equal to $n\cdot(n+1)/2+(n+1)$ which is equal to $(n+1)\cdot(n+2)/2$ by simple calculation. And this is the form we've expected for $n+1$.

Both steps together imply that the statement is true for all $n$.
Counterexample. Now consider the following statement: $1\cdot 2\cdot 3\cdots n=0$. I will use a "shifted" induction here, meaning the induction starts from $n=1$ instead of $n=0$ (which we can do, it is the same thing). Lets assume that we ignore the first step $n=1$ and jump straight to "$n\text{ implies }n+1$" step. Then analogously to the example above we have $1\cdots n\cdot (n+1)=(1\cdots n)\cdot(n+1)=0\cdot (n+1)=0$. This implication is correct, even though the statement is clearly false. That's because we didn't consider the initial $n=1$ step for which $1=0$ statement is false.
