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I think the simplest way to see how it works is as follows:

Given $x=\frac{1}{10n-1}$, it is easy to check that $\frac{x+1}{10}=nx$.

Now $\frac{x+1}{10}$ means shifting the digits to the right, with an $1$ added as the first decimal. For example, in the case of $x=1/19$, we have
$$\frac{x+1}{10} = 0.1\overline{052631578947368421}.$$
But now we see that the additional digit matches the one at the end of the period, so we just can shift the period one to the left. At the same time on the left we use the equation derived above:
$$2\cdot\frac{1}{19} = 0.\overline{105263157894736842}.$$
Thus multiplying by $2$ (or, in the general case, by $n$) is equivalent to shifting the digits one position to the right (and adding a $1$ at the first decimal, which gives the carry from the multiplication). Thus since the last digit of the original period was $1$, the digits to the left of it are successive powers of $2$ (with carry), or successive powers of $n$ in the general case.

So the only remaining question is why the period always ends in $1$. Well, since multiplying the number with $19$ (or, in the general case, with $10n-1 = 10(n-1)+9$) has to give $1=0.\overline{999999999999999999}$, the last digit of the period, when multiplied with the last digit of the denominator, that is $9$, must give $10k+9$ for some integer $k$. But that is clearly only possible if that last digit is $1$.