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I would like to prove that $\forall n \in \mathbb{N},n\geq1$ $, n = 2^k\times m$ with $k \in \mathbb{Z}$ and $ m $ odd and $m \in \mathbb{N}$.

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    Are you allowed to use the prime factorization of $n$ ? This would make it even easier – 2017-01-25
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    This same question has been asked before, cant find it now. – 2017-01-25
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    @Shobhit But we should clarify whether the questions differ by what is allowed to use. Even if the claim is the same, this need not be a duplicate – 2017-01-25
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    The converse of $\forall n \in \mathbb{N}, n = 2^k\times m$ is _not_ $\forall n \in \mathbb{N}, n \neq 2^k\times m$, but rather $\exists n \in \mathbb{N}, n \neq 2^k\times m$ – 2017-01-25
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    I imagine you can remove the idea of contradiction by saying 'if $S$ is non-empty $\ldots$' rather than saying $S$ IS non-empty. Either way you would folllow Martin Gale's hint. – 2017-01-25
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    @Peter i have seen the "exact" same question. – 2017-01-25
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    @Shobhit the question you have seen is http://math.stackexchange.com/questions/1450504/using-the-well-ordering-principle-to-prove-a-property-of-the-integers. but the answers there weren't elaborated on – 2017-01-25

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As a next move, why not consider the cases that $j$ is even or odd? For example, if $j$ is even, you can consider $j'=j/2$ which you know does have the property that $j'=2^k m$ for some $k,m\in \mathbb N$ and $m$ odd, because its smaller than $j$.

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    @IbrahimAbouhashish yes... but for the odd case its even easier to prove the representation you are looking for – 2017-01-25
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Let $a$ and $n$ be positive integers with $a > 2$. Clearly, the nonempty set $\{j \in \mathbb N \text{ s.t } a^j | n\}$ is finite and therefore contains a largest element; call it $k$. Now set $m = \frac{n}{a^k}$, a positive integer. It is clear that is clear that $n = a^km$ and $a$ doesn't divide $n$, by definition of $k$.

The case $a = 2$ is a particular case of the above argument.