How does mathematical induction prove formulas clearly?

The non-circular part is this: induction does not assume the statement is true for every number. It proves two things: 1. The first case is true. 2. If any one case is true, then the next case must be true. Those two facts create a chain. For the formula $$ 1+2+\cdots+n=\frac{n(n+1)}{2}, $$ let $P(n)$ be the statement that the formula is true for that value of $n$. **Base case.** For $n=1$: $$ 1=\frac{1(1+1)}{2}=1. $$ So the first domino is standing. **Inductive step.** Assume the formula is true for $n=k$: $$ 1+2+\cdots+k=\frac{k(k+1)}{2}. $$ Now add the next term, $k+1$, to both sides: $$ \begin{aligned} 1+2+\cdots+k+(k+1) &=\frac{k(k+1)}{2}+(k+1)\\ &=\frac{k(k+1)+2(k+1)}{2}\\ &=\frac{(k+1)(k+2)}{2}. \end{aligned} $$ That is exactly the same formula with $n=k+1$. So we proved: if $P(k)$ is true, then $P(k+1)$ is true. Since $P(1)$ is true, the chain forces $P(2), P(3), P(4)$, and so on. The assumption is temporary. It is not "pretending the theorem is true"; it is proving the logical bridge from one case to the next.