Using the functions $\ell_i$, which are polynomials that depend on the nodes, and based on nodes $x_0,\ldots, x_n$, show that for any $f$, $$\sum_{i=0}^n f(x_i)\ell_i(x)=\sum_{i=0}^n f[x_0,x_1,\dots, x_i]\prod_{j=0}^{i-1}(x-x_j)$$
My Proof: Recall from Lagrange Interpolation Polynomial, $$\ell_i(x)=\prod_{\substack{j=0 \\ j\neq i}}^n\dfrac{x-x_j}{x_i-x_j}.$$ Thus, \begin{equation*} \begin{aligned} \sum_{i=0}^n f(x_i)\ell_i(x) & =\sum_{i=0}^n f(x_i)\prod_{\substack{j=0 \\ j\neq i}}^n\dfrac{x-x_j}{x_i-x_j} \\ & =\sum_{i=0}^n f(x_i)\prod_{\substack{j=0 \\ j\neq i}}^n\dfrac{1}{x_i-x_j}\prod_{\substack{j=0 \\ j\neq i}}^n(x-x_j) \\ & \stackrel{\text{?}}{=}\sum_{i=0}^n f[x_0,x_1,\dots,x_i]\prod_{\substack{j=0 \\ j\neq i}}^n(x-x_j) \\ & \stackrel{\text{?}}{=} \sum_{i=0}^n f[x_0,x_1,\dots,x_i]\prod_{j=0}^{i-1}(x-x_j) \end{aligned} \end{equation*}
So I am unsure of the last two steps can someone help me understand if the logic is correct or do I need to prove each step more?
Your third equality is false. The quantity $f[x_0,...,x_i]$ is the divided difference of $f$ and not the quantity $f(x_i)\prod_{j\neq i} \frac{1}{x_i-x_j}$.
For proving these two formulations (The Lagrange and the Newton ones) are equal, you need to use the unicity of the interpolation polynomial. If you define $P_n(x)=\sum_i f[x_0,...,x_i] \prod_{j=0}^{i-1} (x-x_j)$, you can check that $P_n$ has degree $n$ and $P_n(x_i)=f(x_i)$, so $P_n$ is the Lagrange polynomial. =)