To be a proof, an argument needs to be explicit about the logical structure. An induction proof won't be any different. Take a very standard task like: such as showing that, for all $n\in \mathbb{N}$
$$
\frac{1 - x^{n+1}}{1-x} = 1 + x + x^2 + \cdots + x^n
$$
(i.e., that the rational function on the left is the polynomial on the right).
Answer 1: Let $n\in \mathbb{N}$ be given. We compute
$$
(1 - x)(1 + x + x^2 + \cdots + x^n) =
(1 - x) + (x - x^2) + \cdots + (x^n - x^{n+1})
$$
Rebracketing the rhs we get
$$
1 + (-x + x) + (-x^2 + x^2) + \cdots + (-x^n + x^n) - x^{n+1} = 1 - x^{n+1}
$$
and so
$$
(1 - x)(1 + x + x^2 + \cdots + x^n) = 1 - x^{n+1}
$$
Dividing both sides by $1-x$ gives the identity we wanted for this $n$.
As $n\in \mathbb{N}$ was arbitrary, the identity holds for all $n$.
Comments: I picked this example because it is a standard
school induction exercise, but this isn't a proof by induction.
The logical structure is different. (Though it is the same structure you see in $\varepsilon$-$\delta$ proofs from calculus class. A weird thing to university level instructors is that some students think that "proofs" are "proof by induction" when they arrive, and then struggle with simpler direct proofs like this one.)
Answer 2: The proof is by induction on $n\in \mathbb{N}$. The base case is $n=0$. We verify it directly:
$$
\frac{1 - x^{0 + 1}}{1 - x} = \frac{1 - x}{1 - x} = 1
$$
Now we suppose that, for some $n\ge 0$,
$$
\frac{1 - x^{n+1}}{1-x} = 1 + x + x^2 + \cdots + x^n
$$
Consider now the sum
$$
1 + x + \cdots + x^n + x^{n+1}
$$
By the induction hypothesis, this is equal to
$$
\frac{1 - x^{n+1}}{1-x} + x^{n+1} =
\frac{1 - x^{n+1} + (1 - x)x^{n+1}}{1-x} =
\frac{1 - x^{n+1} + x^{n+1} - x^{n+2}}{1-x} =
\frac{1- x^{(n+1) + 1}}{1-x}
$$
Hence the induction hypothesis implies that the identity holds for $n+1$. This closes the induction, and so we conclude that the identity holds for all $n\in \mathbb{N}$.
Comments: Whether a version of this argument uses the specific template I have (or some other one), it would need to say that induction is being used, what the parameter is, what the IH is and so on. By itself, the final computation is pretty meaningless. Even a very terse version like:
we have $\frac{1 - x^{n+1}}{1-x} + x^{n+1} = \frac{1 - x^{n+2}}{1-x}$ so the identity holds for all $n$ by induction is clear about the logical structure.
You also ask:
If a proof by induction has to explicitly refer to the principle of
mathematical induction, then do proofs using the pigeonhole principle
have to explicitly refer to the pigeonhole principle? Do proofs by
contradiction have to explicitly refer to the principle of proof by
contradiction (if there is such a thing)?
The answer is (obviously?) "yes" to both of these questions. Much like with induction, there is even a standard template for proofs by contradiction, which begins: "Suppose, for a contradiction, that ..."
