Why induction works when the starting point $0$ and the theorem valid for non-zero natural numbers in Peano's arithmetics First of all, I should note that I'm working on Peano's axioms for constructing natural numbers.
For example, lets try to prove that "every natural number is the sum of four squares" by induction.The proof as follows;
i-) Show that $0$ satisfies the condition.
ii-) Assume that $n$ also satisfies the condition, and derive that $n+1$ also satisfies the condition.
The induction should work for proving this kind of a theorem because 0 will satisfies the condition and by applying $ii-)$ where $n=0$, $1$ will also satisfy, and repeating the same procedure, all the natural numbers will satisfy the condition.
However, lets consider the theorem;
"Every non-zero element of N has blah."
In the proof of this kind of a theorem, what it has been done is that they show that $0$ satisfies the theorem (since 0 makes the assumption false), and again apply the second condition of induction.
But why this is also work as it works in the first example ? What is the logic ?
I should also note that, induction is given as the fourth axiom of Peano, so I cannot talk about the proof of induction, but what I am asking is an intuition for the logic why it works.
 A: If you want to prove by induction that every natural number has blah, there are two ways to proceed.  One, suggested in the comments, is to start the induction at $1$ (not at $0$), proving "$1$ has blah" and, for all $n\geq1$, "if $n$ has blah, then so does $n+1$."  
The other approach, which you seem to intend, is to start at $0$ and prove, by induction on $n$, that if $n\neq0$ then $n$ has blah. This statement is, as you observed, vacuously true for $n=0$, so all that remains is to prove, for every natural number $n$, the implication: Assuming "if $n\neq0$ then $n$ has blah" it follows that "if $n+1\neq0$ then $n+1$ has blah." 
That implication can be simplified, but the simplification depends on whether $n=0$. If $n=0$, the induction hypothesis "if $n\neq0$ then $n$ has blah" is vacuously true, so it gives you no real information, and you just have to prove "if $1\neq0$ then $1$ has blah" (because in this case $n+1=1$).  Since in fact $1\neq0$, you need to prove $1$ has blah.  On the other hand, for non-zero $n$, the implication you need to prove simplifies to "if $n$ has blah then so does $n+1$" (because the clauses $n\neq0$ and $n+1\neq0$ are both true in this case).  
Collecting all this information, what the second approach requires you to do is to prove $1$ has blah (to cover the case $n=0$) and to prove, for all non-zero $n$, that if $n$ has blah then so does $n+1$. That's exactly what was required in the first approach, where the induction begins at $1$.
Conclusion: The two approaches look different at first, but, once you analyze and simplify the second approach, it's exactly equivalent to the first.
A: Alternatively, if you want to prove $P(n)$ for $n\in \{1, 2, 3, \cdots \}$, you could instead prove $P(n+1)$ for $n\in \{0, 1, 2, \cdots \}$ by induction. For the base case, prove $P((0)+1)$ or $P(1)$. For the inductive step, you would then, for any $k\in\{0, 1, 2, \cdots \}$, prove that $P(k+1)\implies P((k+1)+1)$ or $P(k+2)$.
