A valid proof by complete induction includes a uniform proof for all $k$ of the inferences below. As such it necessarily includes a proof ($\rm\color{#0a0}{vacuous}$) of the base case $\color{#c00}{\,P(0)}.\,$ See the schema below.
$$\begin{align}
\color{#0a0}{\bbox[3px,border:2px solid #0a0]{\phantom{:}}}\Rightarrow\,\color{#c00}{ P(0)}\\
P(0)\Rightarrow\, P(1)\\
P(0),P(1)\Rightarrow\, P(2)\\
\vdots\qquad\ \ \ \ \\
P(0),P(1),\ldots,P(k-1)\,\Rightarrow\,P(k)\\
\end{align}\qquad\qquad\qquad\qquad\ \ $$
While a valid inductive proof necessarily implies a proof of $\,\color{#c00}{P(0)},\,$ this may not occur explicitly. Rather, it may be a special case of a much more general implication derived in the proof. For example, in many such proofs the natural base case(s) is not a single number but rather a much larger set. Let's examine a simple induction where the base cases are all odd naturals.
If $n\ge\color{#c00}1$ is an integer then $\,n = 2^{\large i} j\, $ for some odd $j$ and some integer $i\ge 0.\,$ For if $n$ is odd then $n = 2^0 n,\,$ else $\,n = 2k\,$ for $\,1 \le k < n\,$ so induction $\,\Rightarrow k = 2^{\large i} j,\,$ so $\, n = 2k = 2^{\large i+1} j.\ \ $ QED
Here the base case $\color{#c00}{P(1)}$ is not explicitly proved. Instead it is a special case of the more general inference that $\,n\,$ odd $\,\Rightarrow\, n = 2^0 n.\,$ In such factorization (decomposition) problems the natural base cases are all irreducibles (and units) - not only the $\rm\color{#c00}{least}$ natural in the statement, e.g. in the proof of existence of prime factorizations of integers $\,n > 1,\,$ with base cases being all primes.
Remark $ $ The same holds for infinite descent - the contrapositive form of complete induction: $\, $ if given a counterexample $\,\lnot P(n)\,$ we can prove there exists a smaller one $\lnot P(k),\ k < n,\,$ then no counterexample exists, else iterating the proof would yield an infinite descending chain of counterexamples, contra $\,\Bbb N\,$ is well-ordered. Or, reformulated, if there is a counterexample then, by well order, we can choose a minimal counterexample (a.k.a. minimal criminal), contra the proof yields a smaller one.
Best Answer
With induction, you can only prove $S(n)$ is true for all positive integers $n$. However, even though $S(n)$ is true for arbitrarily large $n$, the statement "$S(\infty)$" does not follow from induction because $\infty$ is not a positive integer.