Let $\mathbb{N}!$ be the set of permutations of the natural numbers.
With $\aleph_0 = \text{card}\ \mathbb{N}$ you need two facts of cardinal arithmetic to show that the number of permutations of $ \mathbb{N}$ ($\aleph_0! = \text{card}\ \mathbb{N}!$) equals the cardinality of the continuum ($ 2^{\aleph_0} = \text{card}\ \mathbb{R}$):
$\aleph_0\ ! = \aleph_0^{\aleph_0}$
because $\aleph_0 = \aleph_0 -k$ for each $k \in \mathbb{N}$$\aleph_0^{\aleph_0} = 2^{\aleph_0}$
It's easy to see that $\aleph_0^{\aleph_0} \geq 2^{\aleph_0}$ but why is $\aleph_0^{\aleph_0} \leq 2^{\aleph_0}$? (Find the answer below in user egreg's comment.)
I wonder if the following proof – which is more constructive – is valid and interesting. (If it's a well-known proof, please give me a hint, I did not find a reference.) It assumes that you have a representation of the natural numbers (as Cantor did in his diagonalization proofs). For the sake of specificity let's choose the binary representation.
Let $\mathcal{N}$ be the set of normal numbers starting with 1, i.e. $n = 0.1d_2d_3d_4d_5\dots$. It is known that $\text{card}\ \mathcal{N} = \text{card}\ \mathbb{R}$. It is also known that each normal number contains every sequence of digits infinitely often.
Now the proof goes like this:
There is an injective function from $\mathbb{N}!$ to $\mathcal{N}$ which sends each permutation $\pi$ of $\mathbb{N}$ to the unique number $n(\pi) = 0.\pi(1)\pi(10)\pi(11)\pi(100)\pi(101)\dots$ which is normal.
Added: The last claim ("which is normal") is just an assumption, which makes the proof conditional. Even for the identity permutation $\pi_1$ with $\pi_1[k] = k$ and $n(\pi_1) = 0.1\ 10\ 11\ 100\ 101\dots$ it took a proof of its own to show that $n(\pi_1)$ is normal!There is an injective function from $\mathcal{N}$ to $\mathbb{N}!$ which sends each normal number $n = 0.1d_2d_3d_4d_5\dots$ to the unique permutation $\pi(n)$ as explained in this example for $n = 0.10101101011\dots$, written more suggestive as $n = 0.10\ 101\ 1010\ 11\dots$:
Start with the first digit and go to the right until $d_3 = 1$ which yields $\pi(1)=10$.
Do the same with $d_3$: Go to the right until $d_5 = 1$. While the next candidate ($10$) has already been created take one more digit, which yields $\pi(2) = 101$
$\pi(3) = 1010$
$\pi(4) = 11, \dots$.
This way you will eventually find and create all numbers, since normal numbers contain each sequence of digits infinitely often.
So we have
$\text{card}\ \mathbb{N}! \leq \text{card}\ \mathcal{N}$
$\text{card}\ \mathcal{N} \leq \text{card}\ \mathbb{N}!$
$\text{card}\ \mathcal{N} = \text{card}\ \mathbb{N}!$ (by the Schröder-Bernstein theorem)
from which it follows that $\text{card}\ \mathbb{N}! = \text{card}\ \mathbb{R}$.
It's not surprising to see, that you could define a group on $\mathcal{N}$ by $x \circ y = n(\pi(x) \circ \pi(y))$ (if the above proof was correct). Note that the identity element $e$ of this group is the Champernowne constant $C_2 = 0.11011100101110111\dots = 0.862240\dots$. (Note, that it is not obvious that $C_2$ is a normal number: it had to be proved (1992))