Confusing running time analysis for the Divide & Conquer algorithm of Hamiltonian Path problem
Just a note: the $n^{\log n}$ factor can be replaced by a polynomial, without much trouble. Here's one way to do it, not at all optimized...

Imagine there is a global graph $G$ with vertex set $[n]$, and a current induced subgraph $G' = (V',E')$ of $G$, with $n'=|V'|$. Instead of guessing the middle vertex along with each of the ${n' \choose \lfloor n'/2\rfloor}$ subset choices for $S \subseteq V'$, one can recurse only on the choice of $S$. Assume the recursive calls return $n \times n$ matrices $A_S$ and $A_{V'-S}$ such that:

 $A_S[i,j]$ equals the minimum cost of a Hamiltonian tour that starts with some vertex $i \in V$ (not necessarily in $S$), traverses only through vertices in $S$, then ends with some vertex $j \in V$ (not necessarily in $S$).  $A_{V'-S}[i,j]$ is defined similarly.

The algorithm should then return the $n\times n$ matrix $A_{G'}$, where for all $i,j=1,\ldots,n$, $A_{G'}[i,j] = \min_{S \subseteq V'} \min_{k\in V'} (A_S[i,k]+ A_{V'-S}[k,j])$.  Note that we do not have to store all these matrices; we could for example maintain $O(1)$ matrices per recursion layer (and the recursion depth is $O(n)$). After we compute the matrix product $A'[i,j] = \min_k (A_S[i,k]+ A_{V'-S}[k,j])$, we can simply take the MIN of $A'[i,j]$ with the current $A_{G'}[i,j]$ over all $i,j$, then erase $A_S$ and $A_{V'-S}$ before going to the next subset $S$. Thus the space usage is $poly(n)$.

Now the time recurrence is $T(n',n) = {n' \choose \lfloor n'/2\rfloor} \cdot (T(\lfloor n'/2\rfloor,n) + T(\lceil n'/2\rceil,n) + poly(n))$ for all $n' \leq n$ and $T(1,n) = poly(n)$, which yields $T(n',n) \leq 4^{n'} \cdot n^k$ for some $k \geq 1$. To verify this, note that the RHS of the recurrence is upper bounded by

$2^{n'+1} T(n'/2,n) + 2^{n'} \cdot poly(n) \leq 2^{n'+1} \cdot (4^{n'/2} (n/2)^k) + 2^{n'} n^k$
$= 2 \cdot 4^{n'} (n/2)^k + 2^{n'} n^k
\leq 4^{n'} n^k/2^{k-1} + 2^{n'} n^k$,

which is $\leq 4^{n'} n^k$ for $k \geq 3$ (for example).