Why do the factorials appear in differences of consecutive powers?

Question

Why do the factorials appear when repeatedly taking the differences of consecutive powers? Or rather why is the $n_{th}$ factorial equal to the $n_{th}$ difference of $(k+1)^{n}-k^n$? I'm having trouble formulating this comprehensibly, so please see these tables for illustration:

\begin{array}{|c|c|c|c|} \hline 1_{st}.\text{difference} & 2_{nd}.\text{difference} \\ \hline 2^2-1^2 = 3 & \\ \hline 3^2-2^2 = 5 & 5 - 3= 2 \\ \hline 4^2-3^2 = 7 & 7-5=2 \\ \hline 5^2-4^2 = 9 & 9 - 7 = 2 \\ \hline 6^2-5^2 = 11 & 11 - 9 = 2 \\ \hline 7^2-6^2 = 13 & 13 - 11 = 2 \\ \hline \end{array}

\begin{array}{|c|c|c|c|} \hline 1_{st}.\text{difference} & 2_{nd}.\text{difference} & 3_{rd}.\text{difference}\\ \hline 2^3-1^3 = 7 & \\ \hline 3^3-2^3 = 19 & 19 - 7=12 \\ \hline 4^3-3^3 = 37 & 32-19=18 &18-12=6 \\ \hline 5^3-4^3 = 61 & 61 - 37 = 24 &24-18=6 \\ \hline 6^3-5^3 = 91 & 91 - 61 = 30 &30-24=6 \\ \hline 7^3-6^3 = 127 & 127 - 91 = 36 &36-30=6 \\ \hline \end{array}

\begin{array}{|c|c|c|c|} \hline 1_{st}.\text{difference} & 2_{nd}.\text{difference} & 3_{rd}.\text{difference} & 4_{th}.\text{difference} \\ \hline 2^4-1^4 = 15 \\ \hline 3^4-2^4 = 65 & 65 - 15=50 \\ \hline 4^4-3^4 = 175 & 175 - 65 =110 & 110 - 50 =60 \\ \hline 5^4-4^4 = 369 & 369 - 175 = 194 & 194 - 110 = 84 & 84-60=24 \\ \hline 6^4-5^4 = 671 & 671 - 369 = 302 & 302 - 194 = 108 &108-84=24 \\ \hline 7^4-6^4 = 1105 & 1105 - 671 = 434 & 434 - 302 = 132 &132-108=24 \\ \hline \end{array}

\begin{array}{|c|c|c|c|} \hline 1_{st}.\text{difference} & 2_{nd}.\text{difference} & 3_{rd}.\text{difference} & 4_{th}.\text{difference} & 5_{th}.\text{difference}\\ \hline 2^5-1^5 = 31 & \\ \hline 3^5-2^5 = 211 & 211-31=180 & \\ \hline 4^5-3^5 = 781 & 781-211=570 & 570-180= 390 \\ \hline 5^5-4^5 = 2101 &2101 - 781=1320 & 1320-570=750 & 750-390=360 \\ \hline 6^5-5^5 = 4651 &4651 - 2101=2550 & 2550-1320=1230 &1230-750=480 & 480-360=120 \\ \hline 7^5-6^5 = 9031 & 9031-4651=4380 & 4380-2550=1830 &1830-1230=600 & 600-480=120\\ \hline \end{array}

\begin{array}{|c|c|c|c|} \hline 1_{st}.\text{difference} & 2_{nd}.\text{difference} & 3_{rd}.\text{difference} & 4_{th}.\text{difference} & 5_{th}.\text{difference} & 6_{th}.\text{difference}\\ \hline 2^6-1^6 = 63 & \\ \hline 3^6-2^6 = 665 & 665-63=602 & \\ \hline 4^6-3^6 = 3367 & 3367 -665 =2702 & 2702 -602 = 2100\\ \hline 5^6-4^6 = 11529 &11529 - 3367 =8162 & 8162 -2702 =5460& 5460-2100=3360\\ \hline 6^6-5^6 = 31031 &31031 - 11529 =19502‬ & 19502‬ -8162 =11340&11340-5460=5880& 5880-3360=2520\\ \hline 7^6-6^6 = 70993 & 70993 -31031 =39962& 39962-19502‬ =20460&20460-11340=9120& 9120-5880=3240 & 3240 -2520=720\\ \hline \end{array}

Answer 1

For any polynomial of degree $n$, taking $n$ divided differences will get you to a constant. That constant is $n!$ times the coefficient of the leading term. As your polynomial, $k^n$, has a leading coefficient of $1$ the constant will be $n!$.

Answer 2

Let us consider the $n^{th}$ finite difference of powers of $n$ \begin{eqnarray*} \sum_{i=0}^{n} (-1)^i \binom{n}{i} (k+i)^n. \end{eqnarray*} Binomially expand invert the order of the sums & we have \begin{eqnarray*} \sum_{i=0}^{n} (-1)^i \binom{n}{i} \sum_{j=0}^{n} \binom{n}{j} k^{n-j} i^{j} = \sum_{j=0}^{n} \binom{n}{j} k^{n-j} \color{red}{\sum_{i=0}^{n} (-1)^i \binom{n}{i} i^{j}}. \end{eqnarray*} Now the sum in red is zero unless $j=n$ and its value is $n!$, to see this ... observe that \begin{eqnarray*} i^j = [x^j] : e^{ix} j! \end{eqnarray*} and \begin{eqnarray*} \sum_{i=0}^{n} (-1)^i \binom{n}{i} i^{j} &= & [x^j] : \sum_{i=0}^{n} (-1)^i \binom{n}{i} e^{ix} j! \\ &= & [x^j] : (1- e^{x})^n j! \\ \end{eqnarray*}

Answer 3

To make this formulation easier
we need to create our own function to find the nth term of the difference of the sequence (similar to derivative). In case of derivative the function is instantaneous but nth term is only defined for natural number.
Let the new function be $\frac{α}{αn}$. It is defined as
$^1d = \frac{α}{αn}a_n = a_{n+1} - a_n$, 1 in front of $d$ indicates the 1st difference of the sequence.

If $a_n = c$, some constant
Then $^1d=\frac{α}{αn}c=c-c=0$, because it is independent of n ($a_{n+1}=a_n=c$)

2nd difference of the sequence (of $a_n=n^x$),
$^2d=(\frac{α}{αn})^2n^x$
i.e,
$^2d=\frac{α}{αn}(\frac{α}{αn}n^x)$

Now we need to introduce some formulas to make our work easier
\begin{eqnarray*} \frac{α}{αn}n^x&= &(n+1)^x-n^x \\ &= &{x\choose0}n^x+{x\choose1}n^{x-1}+{x\choose2}n^{x-2}+...+{x\choose{x-2}}n^2+{x\choose{x-1}}n+{x\choose x}-n^x \\ &= &{x\choose1}n^{x-1}+{x\choose2}n^{x-2}+...+{x\choose{x-2}}n^2+{x\choose{x-1}}n+1 \\ \end{eqnarray*}$\because$ using binomial theorem

Let's try an example:
$$\frac{α}{αn}n^3={3\choose 1}n^2+{3\choose 2}n+{3\choose 3}$$ $$\tag{1} =3n^2+3n+1 $$ \begin{eqnarray*} \left(\frac{α}{αn}\right)^2 n^3&= &\frac{α}{αn}(3n^2+3n+1) \\ &= &3\frac{α}{αn}n^2+3\frac{α}{αn}n+\frac{α}{αn}1 \\ &= &3(2n+1)+3(1) \\ &= &6n+6 \end{eqnarray*} \begin{eqnarray*} \left(\frac{α}{αn}\right)^3n^3&= &\frac{α}{αn}(6n+6) \\ &= &6=3! \end{eqnarray*}
We finded the third differnce sequence of $n^3$,(i.e, second differnce of $3n^2+3n+1$)
In eq.(1) $3n+1$ have no effect in the result because the they become constant and fade away

so, In finding the kth differnce of any $a_n$

the terms with $n^{th}$ power less than k has no effect on the solution

Now go to our problem \begin{eqnarray*} \left(\frac{α}{αn}\right)^x n^x&= &\left(\frac{α}{αn}\right)^{x-1} \left(\frac{α}{αn} n^x \right) \\ &= &\left(\frac{α}{αn}\right)^{x-1} \left({x\choose 1}n^{x-1}+{x\choose 2}n^{x-2}+...+{x\choose {x-1}}n+{x\choose x} \right) \end{eqnarray*} we are finding ${x-1}^{th}$ differnce of the ${x\choose 1}n^{x-1}+{x\choose 2}n^{x-2}+...+{x\choose {x-1}}n+{x\choose x}$
so terms with less than ${x-1}$ of $n^{th}$ power have no effect hence it reduce to \begin{eqnarray*} \left(\frac{α}{αn}\right)^x n^x&= &\left(\frac{α}{αn}\right)^{x-1} (x n^{x-1}) \\ &= &x\left(\frac{α}{αn}\right)^{x-2} \left( \frac{α}{αn}n^{x-1} \right) \\ &= &x\left(\frac{α}{αn}\right)^{x-2} \left({{x-1}\choose 1}n^{x-2}+{{x-1}\choose 2}n^{x-3}+...+{{x-1}\choose {x-2}}n+{{x-1}\choose {x-1}}\right) \end{eqnarray*} This reduce into \begin{eqnarray*} \left(\frac{α}{αn}\right)^x n^x&= &x\left(\frac{α}{αn}\right)^{x-2} ((x-1) n^{x-2}) \\ &= &x(x-1)\left(\frac{α}{αn}\right)^{x-2} n^{x-2} \\ \end{eqnarray*} continuing we will get \begin{eqnarray*} \left(\frac{α}{αn}\right)^x n^x=x(x-1)(x-2)(x-3)...(3)(2)(1)=x! \end{eqnarray*} comment if any doubt or mistake