You ask: "Is Wikipedia's definition of fractal the standard?" and right near the top of Wikipedia's page of fractals, we see the following definition:
A fractal is a mathematical set that has a fractal dimension that
usually exceeds its topological dimension and may fall between the
integers.
The statement that the fractal dimension may "fall between the integers" really adds nothing but, other than that, I would say that this is fairly standard; it is unquestionably the definition that was put forward by Mandelbrot around 1975 when he coined the term "fractal". He did not refer to "fractal dimension" at that time but, rather, the "Hausdorff-Besicovitch dimension" as he put it. In fairness, the usefulness of this definition has been debated with even Mandelbrot himself feeling that it might not be inclusive enough. Nonetheless, this comparison of dimension is central in fractal geometry. Gerald Edgar calls his great book, Measure, Topology, and Fractal Geometry, a meditation on the definition.
Taking this to be the definition, we can definitely say that the Cantor set satisfies it. If by "fractal dimension" you mean similarity dimension, then the Cantor set has fractal dimension $\log(2)/\log(3)$, since it's composed of two copies of itself scaled by the factor three. Also, the set is regular enough that any reasonable definition of fractal dimension agrees with that computation. (Well, any real-valued defintion.)
Topological dimension is a trickier thing, actually. It's inductive in nature. Totally disconnected sets (like single points, finite sets, or notably the Cantor set) have dimension zero. Higher dimensions are defined in terms of lower dimensions. The space we live in is three dimensions because balls in this space have a surface that is two dimensional. Because of this inductive nature, topological dimension always yields an integer.
When you write that you "do not see the irregular aspects or the complexity that is usually inherent with fractals", I think you might have a bit of a mis-understanding about fractal geometry. The Cantor set is indeed regular but, then so are all the strictly self-similar sets studied in classical fractal geometry - the Koch curve, the Sierpinski triangle, the Menger sponge, and countless others all display this regularity. Indeed, it's exactly this regularity that allows us to understand them.
To emphasize this regularity, and how it appears in not just the Cantor set, compare the following zooms of
The Cantor set
The Koch curve
Now, of course, there are "irregular" fractals - or, at least, less regular fractals. Examples include random version of self-similar sets, examples that arise from number theory, and examples arising from complex dynamics (like Julia sets). It's not their irregularity that makes these objects fractal, however. On the contrary, its the regularity that we can find that allows us to analyse these objects to the point where we can characterize them as fractal. Of course, this analysis is bit harder with these less regular examples.
The 2007 paper you linked in a comment, Bounds of the Hausdorff measure of the Koch curve, contains information which is the most recent I could find. In it, and the related Bounds of Hausdorff measure of the Sierpinski gasket, both by Baoguo Jia, an approach to estimating Hausdorff measures of self-similar sets satisfying the open set condition (citing some books by Falconer on fractal geometry for some facts about the Hausdorff measure for the proof that this method works) is outlined. However, the method is not very effective (it is not easy to calculate good approximations with certainty).
In "Bounds of the Hausdorff measure of the Koch curve", the author proves that the $s$-dimensional (where $s=\log4/\log3$) Hausdorff measure of the Koch curve (base length 1) is bounded below by $$\left(2 \left(\frac{2 \sqrt{3}}{9}\right)^s\right)
\exp\left(-\frac{12 s\sqrt{3}}{9}\right)=2\text{^}\left(-2-\frac{8}{\sqrt{3} \log (3)}+s\right)\approx0.0325239$$and bounded above by $$2 \left(\frac{2 \sqrt{3}}{9}\right)^s=2^{s-2}\approx0.599512\text.$$
At the end of the paper, they conjecture tighter bounds. Assuming their $6/81$ was meant to be ${\sqrt{876}}/{81}$ (the context makes this a reasonable typo), they conjecture a lower bound of $$\left(\frac{1}{122} 4^4
\left(\frac{\sqrt{876}}{81}\right)^s\right)
\exp\left(-\frac{12 s\sqrt{3} }{3^5}\right)=\frac{1}{61}73^{s/2}*2\text{^}\left(s-\frac{8}{27 \sqrt{3} \log
(3)}\right)\approx0.528786$$ and an upper bound of $$\frac{1}{122} 4^4
\left(\frac{\sqrt{876}}{81}\right)^s=\frac{1}{61}73^{s/2}*2^s\approx0.589052\text.$$
In short, if the horizontal base for the curve is $1\mathrm{m}$, and this paper is correct, then the $s$-dimensional Hausdorff measure (which is the appropriate way to measure the size of something a self-similar fractal like this) is definitely less than $0.6\mathrm{m}^{s}$, and probably more than $0.5\mathrm{m}^{s}$.
Best Answer
I want to post the solution, since someone might need it.
Proof. by ordinary induction. Let Induction hypothesis $P(n)$ be $$a_n = a_0\left(\frac{8}{5} - \frac{3}{5}\cdot \left(\frac{4}{9}\right)^n\right).$$ Base case $(n=0):$ $a_0=a_0\left(\frac{8}{5} - \frac{3}{5}\cdot \left(\frac{4}{9}\right)^0\right) = a_0.$ holds.
Inductive step: Assume $P(n)$ holds for some $n \geq 0$. Show that $P(n) \Rightarrow P(n+1).$ We need to show $$a_{n+1} = a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^{n+1}\right).$$ We can write $$a_{n+1} = a_n + e_n t_{n+1}$$ where $e_n = 3\cdot4^n$ and $t_{n+1} = \frac{a_0}{9^{n+1}}.$ Replacing $e_n$ and $t_{n+1}$ in the equation of $a_{n+1}$ gives $$a_{n+1} = a_n + 3\cdot4^n\cdot\left(\frac{a_0}{9^{n+1}}\right)=$$ $$a_n + 3\cdot\left(\frac{1}{9}\right)\cdot a_0\cdot\left(\frac{4^n}{9^{n}}\right)=$$ $$a_n + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ Since P(n) holds, we can replace $a_n$ with $a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^n\right).$ $$a_0\left(\frac{8}{5} - \frac{3}{5}\cdot\left(\frac{4}{9}\right)^n\right) + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot \left(\frac{3}{5}\right)\cdot\left(\frac{4}{9}\right)^n + \frac{1}{3}\cdot a_0\cdot\left(\frac{4}{9}\right)^n=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \left(\frac{3}{5} - \frac{1}{3}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \cdot \left(\frac{4}{15}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^n \cdot \left(\frac{4}{9} \cdot \frac{3}{5}\right)=$$ $$a_0\cdot \left(\frac{8}{5}\right) - a_0\cdot\left(\frac{4}{9}\right)^{n+1} \cdot \frac{3}{5}=$$ $$a_0\cdot \left(\frac{8}{5} - \frac{3}{5} \cdot \left(\frac{4}{9}\right)^{n+1}\right)$$ which proves $P(n+1).$
We can conclude, by induction principle, $\forall n \geq 0: P(n)$.