Is there a closed form for $n^k$ in terms of $\Delta n^{k+1},\Delta n^k$, ...? Let $\Delta$ be a sort of difference operator on a function $f(n)$ such that
$$\Delta f(n)=f(n+1)-f(n)$$
Take the basic power function $f(n)=n^k$, $k\in\mathbb{N}\cup\{0\}$. Then we get
$$\begin{cases}\Delta 1=0\\[1ex]
\Delta n=1\\[1ex]
\Delta n^2=2n+1\\[1ex]
\Delta n^3=3n^2+3n+1
\end{cases}$$
and so on, with the general form
$$\Delta n^k=(n+1)^k-n^k=\sum_{i=0}^{k-1}\binom kin^k$$
I'm wondering if there's a parallel closed form for $n^k$ in terms of $\Delta n^{k+1},\Delta n^k,\ldots,\Delta n$? What I mean by this is that for $n$, we can write
$$n=\frac{\Delta n^2}{2}-\frac{\Delta n}{2}$$
since
$$\frac{\Delta n^2}{2}-\frac{\Delta n}{2}=\frac{2n+1}{2}-\frac{1}{2}=n$$
Similarly, one can show that
$$\begin{cases}n^2=\dfrac{\Delta n^3}{3}-\dfrac{\Delta n^2}{2}+\dfrac{\Delta n}{6}\\[1ex]
n^3=\dfrac{\Delta n^4}{4}-\dfrac{\Delta n^3}{2}+\dfrac{\Delta n^2}{4}\\[1ex]
n^4=\dfrac{\Delta n^5}{5}-\dfrac{\Delta n^4}{2}+\dfrac{\Delta n^3}{3}-\dfrac{\Delta n}{30}\end{cases}$$
and so on, with no immediate pattern as far as I can tell. Is there one?
 A: We  derive  a formula for  $n^k$  in   terms of $\Delta n^k$  by use of binomial inverse pairs and their generating functions. We show

the following is valid
  \begin{align*}
n^k=\frac{1}{k+1}\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}\qquad\qquad k\geq 0
\end{align*}
  with $B_k$ the Bernoulli numbers.

$$ $$

We define
  \begin{align*}
A(x)=\sum_{k=0}^\infty\Delta n^k\frac{x^k}{k!}\qquad\qquad B(x)=\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}
\end{align*}
According to the definition of the $\Delta$ operator we obtain
  \begin{align*}
A(x)&=\sum_{k=0}^{\infty}\Delta n^k\frac{x^k}{k!}\\
&=\sum_{k=0}^{\infty}\left((n+1)^k-n^k\right)\frac{x^k}{k!}\\
&=\sum_{k=0}^{\infty}\sum_{i=0}^{k-1}\binom{k}{i}n^i\frac{x^k}{k!}\\
&=\sum_{k=0}^{\infty}\sum_{i=0}^{k}\binom{k}{i}n^i\frac{x^k}{k!}-\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}\\
&=\left(\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}\right)\left(e^x-1\right)\tag{1}\\
&=B(x)\left(e^x-1\right)\tag{2}
\end{align*}

Comment:


*

*In (1) we use the multiplication of exponential generating functions
\begin{align*}
\left(\sum_{k=0}^\infty a_k \frac{x^k}{k!}\right)\left(\sum_{k=0}^\infty b_k \frac{x^k}{k!}\right)
=\sum_{k=0}^\infty\left(\sum_{i=0}^k\binom{k}{i}a_i b_{k-i}\right)\frac{x^k}{k!}
\end{align*}
with $a_k=n^k$ and $b_k=1$ for all $k\geq 0$.

*In (2) we see the relationship between $\Delta n^k$ und $n^k$ in terms of generating functions.

From (2) we obtain the generating function $B(x)$ for $n^k$ by the generating function $A(x)$ as
  \begin{align*}
B(x)&=A(x)\frac{1}{e^x-1}\\
&=\frac{1}{x}A(x)\frac{x}{e^x-1}\tag{3}
\end{align*}

In (3) we see the generating function of the Bernoulli numbers
\begin{align*}
\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k\frac{x^k}{k!}
\end{align*}

and we obtain
  \begin{align*}
B(x)&=\frac{1}{x}A(x)\frac{x}{e^x-1}\\
&=\frac{1}{x}\left(\sum_{k=0}^{\infty}\Delta n^k\frac{x^k}{k!}\right)\left(\sum_{k=0}^\infty B_k\frac{x^k}{k!}\right)\\
&=\frac{1}{x}\sum_{k=0}^{\infty}\left(\sum_{i=0}^k\binom{k}{i}\Delta n^i B_{k-i}\right)\frac{x^k}{k!}\tag{4}
\end{align*}

$$ $$

Comparing coefficients of $B(x)$ and the RHS of (4) gives
\begin{align*}
\frac{1}{k!}n^k&=\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}\frac{1}{(k+1)!}\qquad\qquad k\geq 0\\
n^k&=\frac{1}{k+1}\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}
\end{align*}
  and the claim follows.

Epilogue Binomial inverse pairs
Let's summarise the result about this binomial inverse pair. We observe the following relationship
\begin{align*}
A(x)&=\sum_{k=0}^\infty\Delta n^k\frac{x^k}{k!}&B(x)&=\sum_{k=0}^{\infty}n^k\frac{x^k}{k!}\\
&=B(x)\left(e^x-1\right)&&=\frac{1}{x}A(x)\frac{x}{e^x-1}\\
\Delta n^k&=\sum_{i=0}^{k-1}\binom{k}{i}n^k&n^k&=\frac{1}{k+1}\sum_{i=0}^{k+1}\binom{k+1}{i}\Delta n^iB_{k+1-i}
\end{align*}
