Is $(\mathbb Z/p^{n+1}\mathbb Z)^\times\cong (\mathbb Z/p\mathbb Z)^\times\times(\mathbb Z/p^n\mathbb Z)$? Let $p$ be an odd prime number and $n$ any positive integer. 
Is $(\mathbb Z/p^{n+1}\mathbb Z)^\times\cong (\mathbb Z/p\mathbb Z)^\times\times(\mathbb Z/p^n\mathbb Z)$ as groups? 
This seems very likely as the orders agree, but I'm unable to find an explicit isomorphism.
Is there one?
Thanks.
 A: The structure of $U(N) = (\mathbb{Z}/N\mathbb{Z})^{\times}$ for any positive integer $N$ is worked out in $\S$ 1.6 of this prebook.  In particular the first theorem in that section affirmatively answers your question.
By the way, your assertion that "this seems very likely as the orders agree" does not seem like a good way of thinking about it to me.  First of all that assertion applies also to $p = 2$, in which case $U(2^a) \cong Z_2 \times Z_{2^{a-2}}$ is not cyclic for all $a \geq 3$.  (Although I suppose a family of statements parameterized by the primes and is true for all $p$ except $2$ is, in a quite reasonable sense, "very likely"!) For any prime $p$, the number of isomorphism classes of commutative groups of order $p^n$ is $P(n)$, the number of partitions of $n$.  Thus two randomly chosen groups of order $p^n$ are very probably not isomorphic.
Now the unit group of a finite ring $\mathfrak{r}$ is not a random finite commutative group.  Given what happens with $\mathbb{Z}/p^a\mathbb{Z}$, you might expect that when $\mathfrak{r}$ has prime power order it is more likely to be cyclic than a random group of its order.  As it happens I did some calculations about a month ago with a large class of unit groups of finite residue rings of Dedekind domains.  I was surprised by how rarely they came out to be cyclic!  This shows up in the proof of Theorem 17 here, which admittedly is the middle of something relatively technical.  A clearer glimpse at the phenomenon might be gained by consulting the short paper of C.S. Dalawat (available on the arxiv) cited therein.
