Since $F \Rightarrow F$ and $F \Rightarrow T$ both evaluate to $T$ with the truth table for $\Rightarrow$, does this not break mathematical induction?
For example, once you show the base case holds for a proposition $P$, then you could do the induction hypothesis as follows: "Suppose $P(k)$ does not hold. Since $P(k+1)$ will either hold or not with this assumption, $P(k) \Rightarrow P(k+1)$, thus $P(n)$ holds for all $n \in \mathbb{N}$."
Mathematical induction works this way :
if (i) $P(0)$ holds and (ii) $\forall n(P(n) \to P(n + 1))$ holds, then we can conclude with (iii) $\forall n P(n).$
In your question, if we have that $P(k)$ does not hold for some $k$, let assume that $k_0$ is the least value of $n$ such that $P(n)$ is false.
This implies that (ii) does not hold, because :
$P(k_0-1) \to P(k_0)$ is false ($T \to F$ is $F$).
No, it doesn't. You can only deduce from true statements. That's independent of the truth table (and even is true in logic systems that don't know truth tables).
But imagine if $F\implies T$ would be false. Then you could prove $1=0$ as follows:
Let's assume $0=1$ were false. Now $0=1 \implies 1=0$ (symmetry). But $1=0\land 0=1\implies 1=1$ (transitivity). Therefore we have proven $0=1\implies 1=1$. But since $1=1$ is undeniably true, that would mean a false statement implies a true statement, violating that $(F\implies T)$ is false. Thus $0=1$ cannot be false by contradiction.
As you can see, actually $(F \implies T)=F$ would break logic.
Your induction hypothesis is missing a case. Certainly, if $P(k) \equiv False$ then you get $P(k) \implies P(k + 1)$. In fact, you get $P(k) \implies Q$ for any proposition $Q$. But what about the case when $P(k) \equiv True$? Here you do not get a trivial $P(k) \implies P(k + 1)$- you must actually do the work to show $P(k + 1)$, assuming the truth of $P(k)$. In fact, a lot of the time, the inductive step is done just this way: assume $P(k)$ to be true and then show $P(k + 1)$ to be true.
