Proving uniqueness of coefficients I have to consider the polynomial $P(x,k)=\sum_{j=0}^{k+1} c_j(k)x^j$ and then get the difference:
$$P(n+1,k)-P(n,k)$$
Which I know is:
$$=c_0(k)+c_1(k)(n+1)+c_2(k)(n+1)^2+\cdots+c_{k+1}(k)(n+1)^{k+1}-(c_{0}(k)+c_{1}(k)n+c_2(k)n^2+\cdots+c_{k+1}(k)n^{k+1})$$
$$=c_0(k)-c_0(k)+c_1(k)(n+1)-c_1(k)n+c_2(k)(n+1)^2-c_2(k)n^2+\cdots+c_{k+1}(k)(n+1)^{k+1}-c_{k+1}(k)n^{k+1}$$
$$=c_0(k)(1-1)+c_1(k)((n+1)-n)+c_2(k)((n+1)^2-n^2)+\cdots+c_{k+1}(k)((n+1)^{k+1}-n^{k+1})$$
$$=\sum_{j=0}^{k+1}c_j \left ( (n+1)^j-n^j \right )$$
I then have to prove that if $\sum_{j=0}^{k+1}c_j\left ( (n+1)^j-n^j \right) =(n+1)^k$ for all positive integers $n,$ then every coefficient $c_j(k)$ is defined in a unique way. I tried to proceed by contradiction, assuming every $c_j(k)$ is equal to a coefficient I'll call $c$, which leaves me with:
$$c(1+\cdots+(n+1)^{k+1}-n^{k+1})=(n+1)^k$$ I feel that must lead me somewhere, but I can't find any contradiciton after that.
 A: The approach with $c$ does not really look promising. Here is an elementary approach by deriving a system of equations in triangle form from which $c_j(k)$ can be iteratively calculated.
We consider the representation with $n,k$ non-negative integers
\begin{align*}
P(n,k)=\sum_{j=1}^{k+1}c_j(k)\left((n+1)^j-n^j\right)=(n+1)^k\tag{1}
\end{align*}
The LHS of (1) is a polynomial in $n$ of degree $k$ since the terms with $n^{k+1}$ cancel out and the RHS is a polynomial of degree $k$.
We denote with $[n^q]$ the coefficient of $n^q$ of a polynomial and make a coefficient comparison of terms with equal exponent in (1).

We start with the RHS of (1) and obtain
\begin{align*}
[n^q](n+1)^k=\binom{k}{q}\qquad\qquad\qquad 0\leq q\leq k\tag{2}
\end{align*}
Now the LHS of (1):
\begin{align*}
[n^q]&\sum_{j=1}^{k+1}c_j(k)\left((n+1)^j-n^j\right)\\
&=\sum_{j=q}^{k+1}c_j(k)\binom{j}{q}-c_q(k)\tag{3}\\
&=\sum_{j=q+1}^{k+1}\binom{j}{q}c_j(k)\tag{4}\quad\qquad\qquad 0\leq q\leq k\\
\end{align*}

Comment:

*

*In (2) we select the coefficient of $n^q$.


*In (3) we select the coefficient of $n^q$.


*In (4) we note that $c_q(k)$ cancels.
Equating (4) and (2) we obtain for $q=k,k-1,\ldots,0$:
\begin{align*}
\begin{array}{l|ll}
q\\
\hline
k&\binom{k+1}{k}\color{blue}{c_{k+1}(k)}\qquad\qquad\qquad\qquad\qquad\quad&=\binom{k}{k}\\
k-1&\binom{k+1}{k-1}c_{k+1}(k)+\binom{k}{k-1}\color{blue}{c_{k}(k)}\qquad\qquad&=\binom{k}{k-1}\\
&\qquad\vdots\qquad\qquad\qquad\qquad\qquad\ddots\\
0&\binom{k+1}{0}c_{k+1}(k)+\binom{k}{0}c_{k}(k)+\cdots+\binom{1}{0}\color{blue}{c_{1}(k)}&=\binom{k}{0}
\end{array}
\end{align*}
Conclusion:

*

*From equation $q=k$ we uniquely derive $c_{k+1}(k)=\frac{1}{k+1}$.


*From equation $q=k-1$ and the knowledge of $c_{k+1}(k)$ we can uniquely derive $c_{k}(k)$.


*We iteratively continue this way from top to bottom till we can finally uniquely derive $c_{1}(k)$.
