$R \cong R[[x]]$ > Is it possible for a commutative ring with 1 to be isomorphic to the power series ring over itself? $$R \cong R[[x]].$$ Bonus marks if you can answer for $n$ variables more generally, $R \cong R[[x_1, \dots, x_n]]$. Obviously, $R$ would have to be huge (infinite Krull dimension). It's not clear to me that taking a colimit of iterates of the functor $S \mapsto S[[x_1, \dots, x_n]]$ produces a ring of the desired kind, essentially because $$R[[x]][[y]] \not \cong R[[x,y]].$$ So you can't make a construction analogous to what's done for $R \cong R[X]$. EDIT: Looks like I was under a misapprehension about the above-stated non-isomorphism. Thank you to several people for correcting this important mistake!

Let $M$ be a monoid which is "locally finite" by which we mean, for all $m\in M$, $\\{(a,b)\in M\times M : ab =m \\}$ is finite.

For such a monoid, for a ring $R$, we can make $R^M$ into a ring with the convolution product instead of $R^{\oplus M}$. Lets denote $R^M$ with the convolution product by $R[[M]]$. Now $R[[x]]=R[[\Bbb{N}]]$. Now observe that as sets, by the usual product-hom adjunction, $(R^M)^N = R^{M\times N}$. Moreover, if $M$ and $N$ are locally finite, then $M\times N$ is locally finite. Indeed, one can check that the preserves sums and products, so $R[[M]][[N]]\cong R[[M\times N]]$. Thus we just need to find a locally finite monoid $M$ such that $M\times \Bbb{N}\cong M$. An easy choice is $$M=\\{u\in \Bbb{N}^\Bbb{N}:u_n\
e 0\text{ for only finitely many $n$}\\}.$$ This is intuitively probably the monoid being referred to when people talk about $R[[x_1,x_2,\ldots]]$.

Thus this gives a proof outline that when $A=R[[x_1,x_2,\ldots]]$, then $A[[y]]\cong A$.