Q:

Why induction is defined as an implication instead of an "if and only if" statement? 

The formal definition of induction, taken from wikipedia, is written as
$$\forall P.\,[[P(0)\land \forall (k\in \mathbb {N} ).\,[P(k)\implies P(k+1)]]\implies \forall (n\in \mathbb {N} ).\,P(n)]$$
But then, it is this version true? 
$$\forall P.\,[[P(0)\land \forall (k\in \mathbb {N} ).\,[P(k)\implies P(k+1)]]\iff \forall (n\in \mathbb {N} ).\,P(n)]\tag{1}$$
If it is not true can you show me why? ( I cant found a counterexample.)  Thank you. 

The background of this question is that if $(1)$ is true then the proof for the equivalence of weak induction ($W$) and strong induction ($S$) is really trivial. 
Let the shortened theorem of strong induction to be $$S\implies \forall (n\in \mathbb {N} ).\,P(n)$$
and by $(1)$ the definition of weak induction $$W\iff \forall (n\in \mathbb {N} ).\,P(n)$$
The case $S\implies W$ is trivial but now the case $W\implies S$ is trivial too because we can verify that 
$$\forall (n\in \mathbb {N} ).\,P(n)\implies S$$

A:

Of course it is true, but the $\Longleftarrow$ direction is trivially true as a matter of logic, so writing $\Longrightarrow$ directs the reader's focus to the interesting direction. 
If $\forall n\, P(n)$ then clearly $P(0)$ is true, and clearly also any formula of the form $\cdots\Rightarrow P(k+1)$ will be true.