 So how does this method work? Well, it turns out that instead of calculating those exponents directly, this time we're going to be doing it recursively. So if I start with an arbitrary number, a, a, b, c, d, e, where each letter represents one digit, maybe the same, maybe not, I want to convert this to something in, say, base 10 from perhaps base 5 again. So the first thing I really want to do is look at what this number means in base 5. So this is a times 5 to the fourth, plus b times 5 cubed, plus c times 5 squared, plus d times 5 to the first, plus e times 5 to the zero. So I could just add all of those terms up if I knew what they were. I'm okay with 5 to the first, 5 squared, even 5 cubed, but getting bigger than that is too much. So I probably don't want to do that arithmetic, especially if I have to do it by hand. But I can do this multiplication kind of piecewise. So I start with nothing. I'm multiplied by 5. I still have nothing. But now I come, I add in a. So now I have a, go back to the beginning, multiply this by 5. Now I have a times 5 to the first. Now I come, I add in b. So let's b. This is the end of the second iteration. Go back to the beginning, start again for the third iteration. I multiply all of this by 5. So now I have b times 5 to the first. And this a times 5 to the first is going to become a times 5 squared. Now I can add in the third term. And I've still got two more terms, so I'll go back and I'll repeat this process again. So I multiply everything by 5. c is now c times 5 to the first. b is now b times 5 squared. a is a times 5 to the third. Second half is adding d. This is the end of the iteration. Go back, start over again. So multiply everything by 5. Now I get d times 5 to the first. c times 5 squared. b times 5 cubed. And a times 5 to the fourth. The second part of the algorithm I add in e. And then since I don't have any more terms, I'm done. So as you can see, these two representations are now the same. But instead of calculating this 5 to the fourth directly, I did it iteratively. Each time I multiplied by my base, until I had enough product of that, that it equaled the expected amount. That's because I multiplied by 5 for each of these terms that was greater than the base 1. Base 1 gets multiplied by 5 to the 0, so that doesn't involve multiplying by 5 at all. But all of these others needed to be multiplied by some number of this base. 1 here, 2 here, 3 here, 4 here. And we just did that iteratively, rather than simply writing them down and calculating it. But as you can imagine, this method is a whole lot easier to do by hand, because we have simple arithmetic to do. We're just multiplying by our base. I don't have to go worry about what is 5 to the fourth, what is 5 cubed. Even worse if I have a huge base that I don't know what to do with, I can just do a whole bunch of multiplication. It's going to be long and ugly, but it's straightforward.