What you mention is not true for all irrational numbers, but for a special subset of them called Normal numbers.
From the wiki article:
While a general proof can be given that almost all real numbers are normal (in the sense that the set of exceptions has Lebesgue measure zero), this proof is not constructive and only very few specific numbers have been shown to be normal.
And
It is widely believed that the (computable) numbers $\sqrt{2}$, $\pi$, and $e$ are normal, but a proof remains elusive.
Note that there are infinitely many irrational numbers that are not normal. In 1909, Borel introduced the concept of a Normal number and proved (with a few gaps resolved later) the following theorem:
Almost all real numbers are normal, in the sense that the set of non-normal irrational numbers has Lebesgue measure zero.
Some additional points of interest: (added with thanks to @leonbloy)
The number of non-normal irrational numbers is uncountable Theorem 4 of this reference.
There is a subset of normal numbers called Abnormal numbers and Absolutely Abnormal numbers which are uncountable. Abnormal numbers are not normal to a given base $b$ while Absolutely Abnormal numbers are not normal to any base $b \ge 2$.
Take an example, say $\frac 1{13}$.
$$
\require{enclose}
\begin{array}{r}
0.0769.. \\[-3pt]
13 \enclose{longdiv}{1.000000} \\[-3pt]
\underline{-91}\phantom{2222} \\[-3pt]
\color{blue}{9}0\phantom{222} \\[-3pt]
\underline{-78}\phantom{211} \\[-3pt]
\color{blue}{12}0\phantom{22} \\[-3pt]
\underline{-117}\phantom{22} \\[-3pt]
\color{blue}{3}0\phantom{2} \\[-3pt]
\vdots\phantom{22}
\end{array}
$$
This sequence $9,12,3$ and so on are the sequence of remainders referred to in the answer. Now, the point is that if the remainder $9$ came again, then doing long division will just repeat the same remainder sequence again (so if you have $9$, you will always bring down the $0$, subtract $78$ and get $12$ as the next remainder, and then $3$ as the remainder after that, and so on).
Note that because each remainder is coming from division by $13$, the remainders are all between $0$ and $12$. Similarly, when we divide by $n$, we get remainders that would be between $0$ and $n-1$.
So, if you want to show that the remainder sequence repeats, then all you need to do, is show that some pair of remainders are the same in the remainder sequence, between $0$ and $n-1$.
Note that if $0$ is a remainder at some point of time , then the long division stops, and there is no repeating part at all (or, depending on which way you look at it, a repeating part of period $1$).
If $0$ is not a remainder, then there are only $n-1$ possible remainders,namely $1,2,...,n-1$. By the $n$th stage, one of these numbers must have occurred twice , since $n>n-1$. However, that shows that the repeating part must come from within the first $n-1$ divisions, so can't be of period more than $n-1$.
Example : keep going with $13$, you eventually get $0.\overline{076923}$, with remainder sequence $9,12,3,4,1,10,9,12,3,...$ where the $9$ repeated by the sixth step, so everything after that repeats as well, giving the repeated decimal.
Best Answer
For $\frac{1}{81}$, there was an $8$, but it got bumped up. We can write $\frac{1}{81}$ as this sum:
$$ \large{\frac{1}{81}}=\;\;\;\small \begin{align} &0.0 \\+\;&0.01 \\+\;&0.002 \\+\;&0.0003 \\+\;&0.00004 \\+\;&0.000005 \\+\;&0.0000006 \\+\;&0.00000007 \\+\;&0.000000008 \\+\;&0.0000000009 \\+\;&0.00000000010 \\+\;&0.000000000011 \\+\;&\underline{\quad\vdots\quad\quad\quad\quad\quad} \\ &0.01234567901 \dots \end{align} $$
This kind of effect of "carrying the $1$" when the nine digit flips to a ten is the thing that is causing the behavior in all of the fractions you are describing.
To follow-up, this should provide a bit more insight as to why interesting patterns appear in the decimal representations of fractions with a power of $9$ or $11$ as the denominator, and see why we can write those numbers like $\frac{1}{81}$ as that sum. First note that $$ \frac{1}{9} = \left( \frac{1}{10}+ \frac{1}{100}+ \frac{1}{1000}+ \dotsb \right) $$ so if we were to consider $\frac{1}{81}$ like before, we would have $$ \frac{1}{81} = \left(\frac{1}{9}\right)^2 = \left( \frac{1}{10}+ \frac{1}{100}+ \frac{1}{1000}+ \dotsb \right)^2 $$ Then if we were to want to know the value of the ten-thousandth's decimal place of $\frac{1}{81}$, we would just have to find the numerator of the term in the expansion of this square with a denominator of $10\,000$, which we can readily see is $$\begin{align} \frac{1}{81} = \Big( \dotsb + \Bigg(\Big(\frac{1}{10}\Big)\Big(\frac{1}{1000}\Big)+\Big(\frac{1}{100}\Big)&\Big(\frac{1}{100}\Big)+\Big(\frac{1}{1000}\Big)\Big(\frac{1}{10}\Big)\Bigg) +\dotsb \Big) \\ = \Big( \dotsb + \Bigg(\frac{3}{10000}&\Bigg) +\dotsb \Big) \end{align}$$
So it looks like these evident patterns that appear in the decimal expansions of fraction with a multiple of $9$ in the denominator is due at least partly to the fact that this infinite sum representation of $\frac{1}{9}$ consists entirely of terms with a numerator of $1$, so multiplying this sum into things may result in "predictable" behavior that results in a pattern.
As for why having a multiple of $11$ in the denominator makes similar patterns, note that $$\begin{align} \frac{1}{11} &= .090909090909090909 \dots \\ &= \left(\frac{9}{100}+\frac{9}{10000}+\dotsb\right) \\ &= \left(\frac{10-1}{100}+\frac{10-1}{10000}+\dotsb\right) \\ &= \left(\frac{1}{10}-\frac{1}{100}+\frac{1}{1000}-\frac{1}{10000}+\dotsb\right) \end{align}$$ So again we have an infinite sum of terms each with a numerator of $1$ (just alternating sign this time) that will result in certain "predictable" patterns when multiplied by things.