Proof of Natural Numbers using n+1 = n ∪ {n} In set theory natural numbers are defined by 0 = ∅ and natural number n+1 = n ∪ {n}
I need to prove that for every n ∈ N , n = {k ∈ N | k < n}.
I know that natural numbers
1 = {∅}
2 = {∅,{∅}} 
3 = {∅,{∅},{∅,{∅}}}
The reason I'm having issues is that my intuition is not even correct. I know we need to use the definition of natural numbers, but I don't understand how n is equal to k ∈ N, when k < n. 
Any help to getting start would be much appreciated!
 A: $n \ne k \in \mathbb N$.  
$n = A;$ some subset of $\mathbb N$ and $k \in A$.
To get you intuition back in tune:
$0 = \emptyset$.
$1 = \emptyset \cup \{0\} = \{0\} = \{\emptyset\}$
$2 = 1 \cup \{1\} = \{\emptyset\} \cup \{1\} = \{\emptyset, 1\} = \{0, \{\emptyset\}\}$.
$3 = 2 \cup \{2\} = \{0, \{\emptyset\}\}\cup \{2\} =  \{0, \{\emptyset\},2\}= \{0, \{\emptyset\},\{0, \{\emptyset\}\}\}$.
And so on.
Or another way of putting it:
$0 = \emptyset$
$1 = 0 \cup \{0\} = \emptyset \cup \{0\} = \{0\}$.
$2 = 1 \cup \{1\} = \{0\}\cup \{1\} = \{0,1\}$.
$3 = 2 \cup \{2\} =  \{0,1\} \cup \{2\} = \{0,1,2\}$
$4 = 3 \cup \{3\} = \{0,1,2\}\cup \{3\} = \{0,1,2,3\}$
And so on.
ANd if we were to replace $1$ with $\{\emptyset,\{\emptyset\}\}$ and so on we'd get the stuff that look more familiar.
Anyway if we use induction and 
suppose $n = \{0,1,2,3,4,......n-1\}$ then
$n+1 = n \cup \{n\} = \{0,1,2,3,4,...., n-1\} \cup \{n\} = \{0,1,2,3,4,.....,n-1, n\}$
And.... that's it.
Think about and reread this until your intuition works.
