Confusion concerning Cantor's theorem. I'm a little confused about Cantor's theorem stating that the cardinality of a set cannot be equal to the cardinality of its power set. 
Consider the power set of $\mathbb{N}$. Couldn't the power set of $\mathbb{N}$ be considered to be a subset of the union of the  sets $\mathbb{N^1}, \mathbb{N^2}, \mathbb{N^3}, \ldots$? All the sets are denumerable, and clearly the list of sets is denumerable, so shouldn't their union and thus an infinite subset of their union be denumerable, implying that the cardinality is the same as $\mathbb{N}$? 
 A: That union gives you only the finite subsets.  There are indeed only countably many finite subsets of $\mathbb N$.
However, we should also note that the union you speak of gives you tuples, not subsets.  For example $(3,1,7)$ and $(1,7,3)$ are two different tuples, as are $(2,2,3)$ and $(2,3)$.  The sets $\{3,1,7\}$ and $\{1,7,3\}$ are the same, and the sets $\{2,2,3\}$ and $\{2,3\}$ are the same.
Cantor's theorem actually says the cardinality of a set is strictly less than the cardinality of its power set.
A: I think there is at work only a small misunderstanding concerning the notion of subset in the sense of infinite subset. I will offer a proof as follows, which I hope helps as it is without explicit reference to infinity.
Theorem: Let $X:=\{x_1,...,x_n\}$ and denote by $P(X)$ the set of all subsets of $X$, then $P(X)>X$. (or equivalently, if $X$ is an $n$-element set, then $P(X)= 2^n$).
Let $f: P(X) \rightarrow D$ when $D := \{s_1,...,s_n\}$, $n \in \mathbb N$.
For each $S \subseteq X$, set $f(S)=(s_1,...,s_n)$ where $s_i=1$ if $x_i$ is in $S$, $0$ if otherwise.
Note that $|D|=2^n$ (or $n$-many copies (products) of elements of $X$).
Proof that $f$ is one to one:
1) Let $S,M \subseteq X$ with $S \neq M$, then $U \nsubseteq V$, so there is some $i, 1<i<n$ such that $x_i$ is in $S$ but not in $M$ (i.e., takes on the value of 1 if in $S$ and $0$ if in $M$. More precisely, for $f(S)$, $f(x_i) =1$ for $x_i \in S$ but $f(x_i) = 0$ for $x_i \in M$. Therefore $ f(S) \neq f(M)$.
Proof that $f$ is onto:
2) Let $(s_1,...,s_n) \in D$ as above. Take $x_i \in S \iff x_i =1$. Clearly, $f(S) = (s_1,...,s_n)$. Thus $f$ is bijective, and $|P(X)|=|D|$, where $|D|$ is $n$-many copies of elements of $X$.
