The "assumption" in proof by induction The second step in proof by induction is to:

Prove that if the statement is true for some integer $n=k$, where
  $k\ge n_0$ then it is also true for the next larger integer, $n=k+1$

My question is about the "if"-statement. Can we just assume that indeed the statement is true? If we assume it, then the proof works... but isn't that similar to the following "proof":

Let $N$ be the largest positive integer.
  Since $1$ is a positive integer, we must have $N\ge1$.
  Since $N^2$ is a positive integer, it cannot exceed the largest positive integer.
  Therefore, $N^2\le N$ and so $N^2-N\le0$.
  Thus, $N(N-1)\le0$ and we must have $N-1\le0$.
  Therefore, $N\le1$. Since also $N\ge1$, we have $N=1$.
  Therefore, $1$ is the largest positive integer.

The only thing that is wrong with this "proof" is that we falsely assume there actually exists a largest positive integer.
So both in the above case and in proof by induction we do an assumption. In the second case the assumption leads to a false conclusion. What is the difference with proof by induction? Why is doing the assumption that the hypothesis is actually true valid here and why doesn't it lead to a similar contradiction?
EDIT: the "proof" above is not mine, it is taken from Calculus a Complete Course 8th edition as an example of why existence proofs are important.

The proof by induction is based on the following statement
\begin{aligned}
& \left [\mathcal {P}(0) \land \left( \mathcal {P}(n) \implies \mathcal {P}(n+1)\; \forall n\geq 0\right)\right] \\
& \implies \mathcal {P}(n)\; \forall n\geq 0 \, ,
\end{aligned}
where $\mathcal {P}$ is a predicate over the natural integers $\Bbb N$.
As soon as one has shown the inheritance property $\mathcal {P}(n) \implies \mathcal {P}(n+1)$ (with the if) and the initialization $\mathcal {P}(0)$, then $\mathcal {P}(n)\; \forall n$. If only the inheritance property is proven, one can obtain such stupid results as all natural integers are larger than $\pi$ (corresponding predicate $\mathcal {P}(n) : n \geq \pi$).
