How many "good" graphs of size $n$ are there? Let's a call a directed simple graph $G$ on $n$ labelled vertices good if every vertex has outdegree 1 and, when considered as if it were undirected, it is connected. How many good graphs of size $n$ are there?
Here's my work so far. Let's call this number $T(n)$. Clearly, $T(2) = 1$: there's only the loop on two vertices.

We also have that $T(3) = 8$. We can count them using the following argument: let's call a possible shape a directed simple graph on $n$ unlabelled vertices which is good. For $n = 3$ we have the following shapes:

There are $3!$ labelled good graphs of the first shape: fix the outside vertex in 3 possible ways, then fix the loop in two possible ways. There are also $\frac{3!}{3}$ labelled good graphs of the second shape: it's simply the number of cycles on 3 elements. So in total we have: $$T(3) = 3! + \frac{3!}{3} = 8\text{.}$$
We also know that $T(4) = 78$. Let's list all possible shapes:

From top left to bottom right, it's easy to check that we have $4!$ labelled good graphs of the first shape, $2\cdot {4 \choose 2}$ of the second, $2\cdot {4 \choose 2}$ of the third, $4!$ of the fourth and $\frac{4!}{4}$ of the last. In total: $$T(4) = 4! + \left(2\cdot {4 \choose 2} + 2\cdot {4 \choose 2}\right) + 4! + \frac{4!}{4} = 3\cdot 4! + \frac{4!}{4}\text{.}$$
I think that $T(5) = 884$, but I won't draw all possible shapes or count their labelings for brevity.
I computed $T(5)$ again, and now I get 944. This also invalidates the following conjecture.
CONJECTURE [DISPROVEN]: I'm conjecturing that there's a simple-ish formula for $T(n)$. It's something like $$T(n) = (2^{n-2} - 1) \cdot n! + \frac{n!}{n} + S(n)$$ where $S(n)$ is some function I currently don't understand such that $S(2) = S(3) = S(4) = 0$, while $S(5) = 5\cdot 4$.
 A: My solution does not agree with your answer for $T(5)$, but let's give it a try anyway . . .
To construct such a graph on $n$ vertices, consider the vertices with indegree $0$.  If there are none of these then the graph is a (directed) cycle, and there are $(n-1)!$ possibilities.  If there are $k$ specified vertices with indegree $0$, then we obtain our graph by choosing a graph of the required type on $n-k$ vertices, then deciding which of these $n-k$ vertices are to be the targets of the edges from the $k$ vertices of indegree $0$.  The number of possibilities is $T(n-k)(n-k)^k$.  Now the maximum value of $k$ is $n-2$ (can't be $n-1$ because then the remaining vertex would have to be adjacent to itself, which you do not allow).  Therefore, by inclusion/exclusion, we have
$$T(n)=(n-1)!+\sum_{k=1}^{n-2}(-1)^{k-1}\!\!\binom nk T(n-k)(n-k)^k\ .$$
The initial value is $T(2)=1$, and for $n=3,4,5$ this gives $8,78,944$.
My results seem to match OEIS A000435, though I can't see any connection with your problem.
Comments please!
A: The graphs you are describing are known as simple (directed) pseudotrees; see http://en.wikipedia.org/wiki/Pseudoforest. There doesn't appear to be a 'nice' closed form for these trees. Wikipedia/OEIS gives the number of undirected connected graphs with $n$ vertices as
$$f(n) =  \sum_{k=1}^n \frac{(-1)^{k-1}}{k} \sum_{n_1+\cdots+n_k=n} \frac{n!}{n_1! \cdots n_k!} \binom{\binom{n_1}{2}+\cdots +\binom{n_k}{2}}{n}.$$
(values at http://oeis.org/A057500). 
Now, this is not quite what you want, since you want to count the number of such directed graphs. Luckily, these two quantities are easily related. 
Note that every pseudotree can be decomposed into a directed cycle and several directed trees attached to points on this cycle. The edges in the directed trees must always point towards the cycle, so their direction is fixed given the undirected the graph. On the other hand, any cycle with more than two vertices can be directed in exactly two ways. Therefore, if $g(n)$ is the number of simple labelled directed pseudotrees (the quantity you want) and if $h(n)$ is the number of  undirected pseudotrees with a cycle of size 2, then $g(n) = 2f(n) - h(n)$.
We can get an expression (although unfortunately not a closed form one) for $h(n)$ by observing that the number of ways to construct an undirected pseudotree with a cycle of size 2 is to choose 2 vertices to belong to the cycle, split the remanining vertices into 2 sets (depending on which vertex they belong to), and count the number of ways to build a labelled rooted forest on each of those sets. By Cayley's formula, there are $(n+1)^{n-1}$ rooted forests on $n$ labelled nodes, so this gives
$$h(n) = \binom{n}{2}\sum_{k=0}^{n-2}\binom{n-2}{k}(k+1)^{k-1}(n-k-1)^{n-k-3}$$
We can in fact generalize this logic to get a sum similar to that for $f(n)$ directly for $g(n)$; the only difference is that if our cycle is of length $k$, we must partition the remaining vertices into $k$ sets instead of 2. This gives
$$f(n) = \sum_{k=2}^{n}\binom{n}{k}(k-1)!\sum_{n_1+\cdots+n_{k}=n-k}\dfrac{(n-k)!}{n_1!\cdots n_k!}(n_1+1)^{n_1-1}\cdots(n_k+1)^{n_k-1}$$
A: Let $f(n)$ be what you want. Denote a graph acceptable if it's connected components (when viewed as an undirected graph) Are all good.(This is the same as saying it is of regular out-degree 1)
Then the number of acceptable graphs is $(n-1)^n$.
But we can also count the number of acceptable graphs by classifying on the number of vertices in the connected component of a special vertex (call it $v$).
We then arrive at the recursion $n(n-1)=\sum_{i=1}^n\binom{n-1}{i-1}f(i)(n-i-1)^{n-i}$.
So $f(n)=(n-1)^n-\sum_{i=1}^{n-1}f(i)\binom{n-1}{i-1}(n-i-1)^{n-i}$
