Finding an error bound for Lagrange interpolation with evenly spaced nodes

Question

I know that the error bound for Lagrange interpolation is usually $$\frac{M_{n+1}}{(n+1)!}\max_{x\in[a,b]}|(x-x_0)\cdots(x-x_n)|$$, where $M_i=\max_{x\in[a,b]}|f^{(i)}(x)|$. I'm trying to find the error for when the points upon which we are interpolating are evenly spaced, so we write $x=x_0+(k+t)h$, where $h=\frac{b-a}{n}$, $k=0,\dots,n-1$ and $0\leq t<1$. I have some hints to guide me, and it says the first step to finding the error is showing that $$|(x-x_0)\cdots(x-x_n)|=h^{n+1}(k+t)(k-1+t)\cdots(1+t)t(1-t)(2-t)\cdots(n-k-t)$$ So far, I figured out this up until the term $t$. What I've figured so far is that by the above formula, $(x-x_0)=(k+t)h$, and then for each subsequent $x_i$, since they are one more interval $h$ from x, we subtract $nh$. This has given me $|(x-x_0)\cdots(x-x_n)|=[h(k+t)][h(k+t)-h]\cdots[h(k+t)-nh]=h^{n+1}(k+t)(k+t-1)\cdots(k+t-n)$. However, judging by the hint given above this is evidently wrong and I'm wondering where I went wrong, mainly finding out where the product ends, and why it contains the factors $(1+t),t,(1-t),(2-t)$ in particular.