Counterexample for $\prod_{i<\nu} \kappa_i=(\sup_{i<\nu}\kappa_i)^{\nu}$ when $\kappa_i$ is not an increasing $\nu$-sequence of cardinals

Question

It is known that $\displaystyle \prod_{i<\nu} \kappa_i=\left(\sup_{i<\nu} \kappa_i \right)^{\nu}$ if $\nu$ is an infinite cardinal and $\langle \kappa_i | i < \nu \rangle$ is an increasing $\nu$-sequence of infinite cardinals. I'm stuck at finding a counterexample if the increasing monotony hypothesis doesn't hold. I tried writing $\nu=\nu \times \nu$ or using $cof(\nu)$ in some ways, but the most that I have achieved is that I should obtain $\displaystyle \prod_{i<\nu} \kappa_i<\left(\sup_{i<\nu} \kappa_i \right)^{\nu}$ cause the inequality with $\le$ is always true.

Answer 1

Take $\lambda$ to be some cardinal such that:

1. $\lambda>(\sup\kappa_i)^\nu$,
2. $\operatorname{cf}(\lambda)\leq\nu$.

Now define $\lambda_0=\lambda$ and $\lambda_{1+\alpha}=\kappa_\alpha$ for all $\alpha<\nu$. Then we have $$\prod_{i<\nu}\lambda_i = \lambda\times\prod_{i<\nu}\kappa_i=\lambda\times(\sup\kappa_i)^\nu=\lambda<\lambda^{\operatorname{cf}(\lambda)}\leq\lambda^\nu=(\sup\lambda_i)^\nu.$$