For $n>1$, I want to find the smallest value, and corresponding $x_i$ values, of $f(x_2,\dots,x_n) = \prod_{k=2}^n (x_k+1)^k$ subject to the constraints $x_j > 0$ for all $j$ and $\prod_{k=2}^n x_k = 1$.
I've been trying to find a solution with Lagrange multipliers but I'm struggling with the fact that each $x_k$ depends on each other. This is not homework; now fixed the typo.
Apply Lagrange's method to $$g(x):=\log\bigl(f(x)\bigr)=\sum_{k=2}^n k\log(1+x_k)$$ under the given constraint. This involves setting up the function $$\Phi(x,\lambda):=g(x)-\lambda\prod_{k=2}^n x_k$$ and solving the system $${\partial\Phi\over\partial x_k}=0\quad(2\leq k\leq n),\qquad \prod_{k=2}^n x_k=1\ .$$ This gives $${k\over 1+x_k}={\lambda\over x_k}\qquad(2\leq k\leq n)\tag{1}$$ and therefore $$x_k={\lambda\over k-\lambda}\ ,$$ where $\lambda$ has to be determined such that $\prod_{k=2}^n x_k=1$. From $(1)$ it then follows that $$\prod_{k=2}^n(1+x_k)={n!\over\lambda^{n-1}}\ .$$ (Note that we haven't yet discussed the question whether we actually have a minimum of $f$ at the point $x$ found in this way.)