Given a sequence $y_0,y_1,...,$ define the forward difference operator $\Delta$ by $$\Delta y_i = y_{i+1}-y_i$$ Powers of $\Delta$ are defined recursively by $$\Delta^0y_i=y_i,$$ $$\Delta^jy_i=\Delta(\Delta^{j-1}y_i), \hspace{10mm} j=1,2,...$$ Consider a polynomial interpolation at equispaced points, $x_i=x_0+ih, \hspace{4mm}i = 0,1,...,n$
Show that $$f[x_0,x_1,...,x_j]=\frac{1}{j!h^j}\Delta^jf(x_0)$$ using mathematical induction
My attempt:
Checking the base case is pretty straightforward, so I will skip that. My issue is proving for $j+1$.
By using $f[x_0,....,x_j]=\frac{f[x_1,...x_j]-f[x_0,...,x_{j-1}]}{x_j-x_0}$ and rewriting using $j+1$ I get: $$f[x_1,....,x_{j+1}]-(x_{j+1}-x_0)f[x_0,....,x_{j+1}]=f[x_0,...,x_j]$$ We assume $P(j)$ is true and we have $x_{j+1}-x_0=(j+1)h$, so we get $$f[x_1,....,x_{j+1}]-((j+1)h)f[x_0,....,x_{j+1}]=\frac{1}{j!h^j\Delta^jf(x_0)}$$ So I think I am on the right track here or getting close, but I am stuck on how to handle $f[x_1,....,x_{j+1}]$. My intuition is telling me that I could conclude something due to the indices being shifted by 1, but I am not quite sure what that is. Any help is appreciated, thank you.