Let us assume we already know the Principle of (weak) Mathematical Induction (if $\mathrm{P}(0)$ is true and $\mathrm{P}(n+1)$ is also true whenever $\mathrm{P}(n)$ is, then $\forall n \in \mathbb{N} \ \ \mathrm{P}(n)$ is true) which is really a trivial consequence of the construction of $\mathbb{N}$. We can also assume the well-ordering principle for $\mathbb{N}$.
What I need to prove is the following "generalized" (weak) induction principle:
$\forall k \in \mathbb{N} \ \ \forall$ properties $\mathrm{P}(x)$
($\mathrm{P}(k)$ holds and $\forall n \in \mathbb{N}, k \leq n: \ \mathrm{P}(n) \Rightarrow \mathrm{P}(n+1)) \Rightarrow \forall n \in \mathbb{N}, k \leq n: \ \mathrm{P}(n)$ holds
I have tried to use the ordinary induction on $k$. For $k = 0$ it becomes the Principle of (weak) Mathematical Induction.
However, even assuming it's true for $k$, I can't seem to prove it also is for $k+1$.
Before asking this question, I was sure it had been asked already, but I can't seem to find a prood of that exact theorem. It's either the weak induction with a basis $0$, or the strong induction.
Sure, fix $k$. Define the property $Q(n)$ as $P(k+n)$.
Now we can prove that $Q(0)$ holds, and $Q(n)\rightarrow Q(n+1)$. By induction, we get that for all $n$, $Q(n)$ holds. Which translates to $P(n)$ for all $n\geq k$ holds.
Alternatively, define $Q'(n)$ as "true" for $n<k$ and $P(n)$ for $n\geq k$, and again prove by induction that $\forall n: Q'(n)$ holds.
Since our $k$ and $P$ were arbitrary, the conclusion follows.