https://www.youtube.com/watch?v=rEoYwyHddLc
Explaining why it works is an exercise in number theory. For example, card 1 stays in place; card 2 goes to position 3, then 5, then 9, then 17, ... In short, the reason why it works is that 2^8 - 1 is divisible by 52 - 1.
The in shuffle is n shuffles in general to get back to the start but the out shuffle isn't such a simple pattern.
If you want to make a shuffle that will take _even longer_ to get back to it's starting point, the best you can do is the Landau function which as you can see can get very big: https://oeis.org/A000793/list (I have a calculation of g(52) but not on me right now)
http://statweb.stanford.edu/~cgates/PERSI/papers/83_05_shuff...
I saw one of the authors, Persi Diaconis, give a talk on this paper, and then perform a perfect shuffle. I was floored.
Also, the statement in the article is for a random riffle, not a perfect one. Any deck which has only been perfectly riffled is entirely determined.
Of course, this is dependent on perfect shuffles, which I'm sure I never achieve. Maybe the 'seven shuffles to randomize' calculation takes into account the 'human' nature of shuffling during a game? It is almost a sure thing the deck will never be split perfectly in half, and a perfect faro shuffle achieved.
Someone else mentioned that it depends on if it's an 'in' or an 'out' riffle as well, so I read the wiki page. The basically tells me you should always try to do an 'in' shuffle, I will have to start looking out for this :)