What I am talking about is a fact, that if we write down n-th powers of consecutive natural numbers in a row, and then on the next row between each two numbers write their difference and repeat this algorithm, then on the n-th row we always get a row of $n!$'s. So why it is like that? The only way I can associate power with factorial is series, but I think it's not the case. Here is an example of the thing I am talking about:

1       32      243     1024    3125    7776    
    31      211     781     2101    4651    
        180     570     1320    2550    
            390     750     1230    
                360     480

$n$-th (last) row (if we assume that first one is $0$th) is always $n!$ (in this case $5!$).

I hope my question is clear. Thanks in advance!


  • $\begingroup$ Can you perhaps explain how the numbers above were computer? in particular, how was the first row computed? $\endgroup$ Oct 16, 2012 at 20:55
  • $\begingroup$ thanks, I approved your edit. each number in row $n\geq 1$ is a difference of two numbers standing above. $\endgroup$
    – nakajuice
    Oct 16, 2012 at 21:01

2 Answers 2


This is well known and very useful. My father taught me the method of finite differences when I was a child, and I found your case by experiment. The main explanation is induction in two directions... If I write row $0$ as a polynomial of degree $n,$ inyour case $n=5$ and the entries are $x^5,$ then row $1$ is a polynomial of degree $n-1,$ in this case $n-1 = 4$ and $$ (x+1)^5 - x^5 = 5 x^4 + 10 x^3 + 10 x^2 + 5 x + 1. $$ The next row is, separately and then summed, $$ 5((x+1)^4 - x^4) + 10((x+1)^3 - x^3) + 10((x+1)^2 - x^2) + 5((x+1)- x) + (1-1) $$ so the result is that row $2$ has degree $3.$ Meanwhile, the coefficient of the highest degree term in each row occur as $x^5, \; 5 x^4, \; 20 x^3, \; 60 x^2, \; 120 x, \; 120.$ If you continued another row you would get all $0$'s. So this is very similar to taking ordinary calculus derivatives. Furthermore, if you begin with integer coefficients, you continue to get integer coefficients in later rows. Meanwhile, note that i have been a little careless about $x.$ In each row, $x=1$ should be the first entry, so that location goes on a diagonal, sort of backwards of what Pascal's Triangle does.

Finally, this is a guaranteed method for detecting polynomials. If you begin with a string of numbers and do this procedure, eventually you get all $0$'s. Why don't you start with a polynomial such as $5 x^3 + 7 x^2 + 11 x + 13 $ and do the same procedure? Oh, you always take the differences in the same order, even if that means the result is zero or negative. If you start throwing in absolute values this all goes wrong.

Here is one LINK


By the binomial development, $$(n+1)^k-n^k=kn^{k-1}+\text{low order terms in }n.$$

If you iterate the differencing, the leading terms of these polynomials evolve like $k(k-1)n^{k-2},k(k-1)(k-2)n^{k-3},\cdots$ and in the last row the constant $k!n^0$.



$$4(n+1)^3+6(n+1)^2+4(n+1)+1-4n^3-6n^2-4n-1=\color{green}{12}n^2+24n+14,$$ $${12}(n+1)^2+24(n+1)+14-{12}n^2-24n-14=\color{green}{24}n+36,$$ $$24(n+1)+36-24n-36=\color{green}{24}.$$


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