Why does induction need a base case if the step works?

The first case plus the step produces a finite route to every whole number. From $P(1)$, the step with $n=1$ gives $P(2)$; the step with $n=2$ gives $P(3)$; and so on. Check the arithmetic on the first two links: $P(2)$ says $1+2=2\cdot3/2$, both sides $3$; $P(3)$ says $1+2+3=3\cdot4/2$, both sides $6$. The step is the same general argument each time, not a new check for $2$, $3$ or any later number. Without $P(1)$, the first application has no input.

People sometimes read the step as a licence to check $P(1)$, then $P(2)$, then $P(3)$, and so on forever. It is not that. The step is one general implication, proved once for an arbitrary $n$, and the base case is the one place where the condition of that implication is met outright. That is why two finite checks settle infinitely many cases.

Here is a claim with a step that works for every $n$ and is false for every $n$: $P(n)$ is $n=n+1$. The step says: if $n=n+1$, then adding $1$ to both sides gives $n+1=n+2$, which is $P(n+1)$. That conditional is true for every $n$. The base case fails at $n=1$, since $1\ne2$. Nothing follows, and indeed no number equals the next. This is the plainest case of a perfect step proving nothing: the step is true, the claim is false everywhere, and the missing base is the whole difference.

The natural guess is that a step covering every transition must cover every case. It does not. A transition is not a case. The step is a conditional, and a conditional can be true while its antecedent is false for every number; $n=n+1$ is exactly that. The step never asserts $P(n)$ for any $n$, so it cannot manufacture $P(1)$. The base case is the only place a case is asserted outright. Remove it and the argument has no true instance to apply the step to.

A step can also fail to work for the general case even though it works for the small cases you tried. The classic shape is the claim that every group of $n$ horses is one colour. The first case is true: a group of one horse has one colour. The step from $n$ to $n+1$ is argued by taking two overlapping groups of $n$ horses, each one colour, and concluding the union is one colour. That argument needs the two groups to overlap. For $n=1$, each group has one horse, and two groups of one horse cannot overlap unless they are the same horse, so the step from $1$ to $2$ is not established. The step works only for $n\ge2$, and the chain never reaches $2$.

Another step may need more than the single case before it. If it uses $P(n-1)$ as well as $P(n)$, or several earlier cases, the single-step form is invalid. It can be repaired by a strong step that assumes all earlier cases, but the repair must be stated and the base case still has to start the chain. A step that quietly uses two previous cases is a step nobody can check from the written argument. The first case and the step are two separate checks, and neither replaces the other.