system
Given the following post, respond as if you're commenting on Reddit:
human
Does this* number have a name?

I would be more specific in the title, but it would be too incomprehensible and cut-short probably. It will take some explanation before I get to my question again but it's there:

I was working on a math puzzle a while back where I had to guess the next number in a series. The given series was "1, 4, 9, 16, 25, 36, 49, __" and I had to guess the last number. I guessed correctly, but instead of intuitively realizing that the pattern was " x^2 ", I found the answer via another method.

I found the differences between the adjacent numbers in the series, i.e. 4 - 1 = 3, 9 - 4 = 5, so on, and used those to create a new series: "3, 5, 7, 9, 11, 13". I then performed the same operation on that line, finding the differences between adjacent numbers and then using those to create a new line: "2, 2, 2, 2, 2".

I thought this was interesting, because by performing this differences method twice on my given series, I was able to bottom-out/hit a ceiling at the number 2*. The number 2 was useful, because I could then work my way back down/up (depends on how you write it down, I guess) to find the next number in the series: 64.

I decided to try this again on a new, related series. I made the series "1, 8, 27, 64, 125" ( x^3 ) and resolved to use the same differences method on this line. The first time yields: 7, 19, 37, 61"; the second time yields: "12, 18, 24"; the third time bottoms out at: "6, 6".

So ( x^2 ) bottoms out at 2*, and ( x^3 ) bottoms out at 6*.

I continued this with ( x^4 ): "1, 16, 81, 256, 625, 1296, 2401". The differences method yields: "15, 65, 175, 369, 671, 1105"; again, and it yields: "50, 110, 194, 302, 434"; again, and it yields: "60, 84, 108, 132"; again, and it yields: "24, 24, 24".

So ( x^4 ) bottoms out at 24*.

Finally, my question: Is there a name for these bedrock numbers that I've marked with the asterisks? 

Additionally, I would like to know: is there reading material on the matter? Is there a more elegant way of finding the bedrock number that underlies a given exponent?

tl;dr:

2 underlies ( x^2 ),
6 underlies ( x^3 ),
24 underlies ( x^4 ),
and so on, for reasons explained above.
Why?
gpt
>Is there a more elegant way of finding the bedrock number that underlies a given exponent?

Yes, for x^n the number you're looking for will be n!, which is shorthand for n \* (n-1) \* (n-2) \*....\*3\*2\*1.

For example, x^7 would end up with the number 7! = 7\*6\*5\*4\*3\*2\*1 = 5040

To see why:
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Focus on x^2 to start. 

To generalize the difference between consecutive items in the sequence, we have:

(x+1)^2 - x^2

Do some algebra and get:  
**2**x^**1** + 1

Repeat the process:

[2(x+1) + 1] - [2x+1]  
= 2

So there's your 2.

--------
Now we do this with x^3:

(x+1)^3 - x^3  
= **3**x^**2** + 3x + 1

Repeat the process:  
[3(x+1)^2 +3(x+1) + 1] - [3x^2 + 3x + 1]  
= **6**x^**1** + 6 

Repeat the process:  
6(x+1) + 6 - [6x+6]  
= 6 [aka **6**x^**0**]

Making sense of it:
-------

Notice that I've bolded a few coefficients and exponents in the above section. As you go through the repetitive process, you see that your highest power term always gets canceled out in the algebra. In the x^3 example, you start with 1x^3 being your highest power term. Then you do the difference process and you end up with 3x^2 being your highest power. Then you do the difference process again and you end up with 6x^1 being your highest power. Then you do it once more and 6x^0 is your highest power. 

**In the x^3 example, do you see how the highest power term goes like this as we go through all the steps:**

x^3  
3x^2  
6x^1  
6x^0

**So the general pattern is:**  
x^n  
nx^(n-1)  
n(n-1)x^(n-2)  
n(n-1)(n-2)x^(n-3)  
.  
.  
.  
n(n-1)(n-2)...3\*2\*1

which is the formula we talked about up top.