$COF(\lambda)$ is stationary in $k$, where $\lambda < k$ is regular.

Question

I want to prove that:
if $\lambda < k$ is regular then $COF(\lambda) = \{\alpha < k | cof(\alpha) = \lambda \}$ is stationary in $k$.
I think I have to use that if $\alpha \neq \lambda$ is a cardinal in $COF(\lambda)$ then $\alpha$ is singular, so exist an increasing sequence $<k_i | i < cof(\alpha) = \lambda>$ such that $\alpha = sup_{i < \lambda} k_i$.
If there's a way to show that i can choose the $k_i$-s in a club $C$ for every club C of $k$ then the proof is done. How can I make it? Is the idea correct? Maybe something is missing because I don't see where I would use the regularity of $\lambda$. Any advice?

Answer 1

Consider any club subset of $\kappa $. Check that it has order type $\kappa>\lambda $, and that its $\lambda $th element (in its increasing enumeration) has cofinality $\lambda $.