Let's orient a $K_5$ to have arcs 12, 23, 34, 45, 51 (which I think is enough to make it strongly connected), 13, 14, 24, 25, 35. Each vertex has total degree $4$, which is even, but vertex $1$ has indegree $1$ and outdegree $3$.
Tarjan's algorithm may be used to detect cycles in a directed graph as follows: When the root node of a (maximal) strongly connected component (SCC) is detected, the nodes in the SCC are popped from the stack. At that point, if more than one node is popped, then it is known that the graph contains a cycle. If the SCC contains a single node, one has to check whether that node has a "self-loop" (which gives a cycle) or not (that is, the node makes up a trivial SCC).
Said otherwise, if the directed graph has no cycles, Tarjan's algorithm will only find SCCs consisting of individual nodes without self-loops.
If the cycle needs to be identified and the SCC found has more than one node, one can perform a depth-first search from the SCC root, limited to the nodes that are in that SCC and until the root of the SCC is re-entered.
In fact, in practice, cycle detection in directed graphs is performed by two nested depth-first searches, without concern for the enumeration of SCCs, especially when cycles have to go through some designated nodes.
When the "outer" DFS is about to retreat from a (designated) node $v$, a second DFS is run from $v$ to see if some node $u$ currently on the stack of the outer DFS may be reached from $v$.
If that is the case, we have a cycle reachable from the node from which the first DFS started. Such a cycle goes through both $v$, the node from which the outer DFS is about to retreat, and $u$, the node on the stack of the outer DFS found during the inner DFS. We know we can reach $v$ from $u$ because $u$ is on the stack of the outer DFS. We know we can reach $u$ from $v$ because the inner DFS says so.
The nested DFS algorithm is not asymptotically faster than Tarjan's algorithm, but often performs better.
Best Answer
We can prove this by induction on $n$. For $n=3$, it is clear that the only strongly connected digraph is the $3$-cycle. Now suppose for some $n\geqslant 3$ that the only strongly connected digraph on $n$ vertices is the $n$-cycle, denoted $C_n$. Adding a vertex $v$, we see that in order for $v$ to have indegree and outdegree $1$, there must be vertices $u,w\in C_n$ such that $uv$ and $vw$ are edges in $C_n\cup\{v\}$.
Now, if we add simply add these edges to the graph, then $u$ will have indegree $2$ and $w$ will have outdegree $2$, which is not acceptable. However, if $u$ and $w$ are adjacent, that is, $uw$ is an edge in $C_n$, then we can remove the edge $uw$ and then add the edges $uv$ and $vw$ and obtain $C_n\cup\{v,uv,vw\}\setminus\{uw\}=C_{n+1}$.
It perhaps remains to show that there is no other method of obtaining a strongly connected digraph with $n+1$ vertices from a strongly connected digraph with $n$ vertices. But I will leave that proof to you, if you are convinced it is necessary.