Since the link is broken, here is my proof. Let me know if it would helpful to include any further details anywhere, or if you can shorten any parts of the proof.
This question was also discussed in this MO post.
Let $G$ be simple of order $720 = 16 \times 9 \times 5$.
By Sylow, $|{\rm Syl}_3(G)| = 1, 4, 16, 10$ or $40$.
It is clearly not $1$ or $4$.
By Sylow, all groups of order $45$ are abelian, so $|{\rm Syl}_3(G)|$ cannot be
$16$ by BTT (Burnside's Transfer Theorem).
We need to eliminate $|{\rm Syl}_3(G)| = 40$.
If $|{\rm Syl}_3(G)| = 40$, then $P \in {\rm Syl}_3(G)$ has an orbit of length
$3$ in the conjugation action of $G$ on ${\rm Syl}_3(G)$, so
there is a subgroup $Q$ of order $3$ (the pointwise stabilizer of this orbit
in $P$) such that $N := N_G(Q)$ has more than one Sylow $3$-subgroup.
So it has at least four, and we get $|N| = 36$ or $72$.
If $|N| = 36$ then $N/Q$ has order $12$ and has four Sylow $3$-subgroups, so
$N/Q = A_4$, and since $A_4$ cannot act non-trivially on $Q$, $Q$ is central
in $N$. Hence $N$ has a normal subgroup $T$ of order $4$, and $|N_G(T)|$ is
divisible by $8$, so strictly contains $N$. Then the only possibility is
$|N_G(T)| = 72$, but then $Q = O_3(N)$ is characteristic in $N$ and hence
normal in $N_G(T)$, contradiction, since $N = N_G(Q)$.
So $|N| = 72$. Since $|{\rm Aut}(Q)| = 2$, $C(Q)$ has order at least $36$, and a
subgroup $R$ of order $12$ in $C(Q)$ must be abelian.
Consider the action of $G$ on the $10$ cosets of $N$. The subgroup $N$ must be maximal in $G$ because any larger subgroup would have too small an index. So this action is primitive. Let $Q = \langle t \rangle$.
By the result of Jordan proved below, $t$ cannot consist of a single $3$-cycle.
If $t$ consists of two $3$-cycles,
then an element $u$ of order $2$ in $R$ must interchange those cycles forming
a $6$-cycle $tu$. Since the $6$-cycle is self-centralizing in $S_6$,
an element in $R$ outside of $\langle tu \rangle$ must fix all $6$ points of
the $6$-cycle, so it must be a single transposition, which is impossible.
If $t$ consists of three $3$-cycles, then an element of order $2$ in $R$ must interchange
two of these $3$-cycles and fix the other pointwise, so it consists of three
$2$-cycles, and is an odd permutation, which is impossible in a simple group.
So $|{\rm Syl}_3(G)| = 10$. Let $P \in {\rm Syl}_3(G)$ and $N = N_G(P)$, so
$|N| = 72$ and $G$ acts transitively by conjugation on ${\rm Syl}_3(G)$,
which we denote by $\{ 1,2,\ldots,10 \}$,
with $P = 1$, and $N = G_1$ the stabilizer of $1$ in $G$.
If $P$ is cyclic then it must act as a $9$-cycle on $\{ 2,\ldots,10 \}$.
Since $|{\rm Aut}(P)| = 6$,
there is an element of order $2$ in $N$ which centralizes $P$, and there is no
way for such an element to act on $\{ 2,\ldots,10 \}$.
So $P$ is elementary abelian. If a subgroup $Q$ of $P$ of order $3$ fixes more
than one point, then $N_G(Q)$ has more than one Sylow $3$-subgroup,
giving a contradiction as before.
So $P$ acts fixed-point-freely on $\{ 2,\ldots,10 \}$. In fact we can assume
that $P = \langle a,b \rangle$ with
$$a = (2,3,4)(5,6,7)(8,9,10),\
b = (2,5,8)(3,6,9)(4,7,10).$$
The stabilizer $S = N_2$ of $2$ in $N$ has order $8$ and is a Sylow
$2$-subgroup of $N$. Now $S$ is contained in $X_2$, where $X$ is the
normalizer of $P$ in the symmetric group on $\{ 2,\ldots,10 \}$,
and $X_2$ can be identified with ${\rm Aut}(P) = {\rm GL}(2,3)$.
Note that the element $(5,8)(6,9)(7,10)$ of $X_2$ is an odd permutation and
corresponds to an element of determinant $-1$ in ${\rm GL}(2,3)$.
Since ${\rm SL}(2,3)$ is the unique subgroup of index $2$ in ${\rm GL}(2,3)$,
it follows that the elements of ${\rm SL}(2,3)$ correspond to the even
permutations in $X_2$. So $S$ corresponds to a Sylow $2$-subgroup of
${\rm SL}(2,3)$, which is isomorphic to $Q_8$. In fact ${\rm SL}(2,3)$ has a
unique Sylow $2$-subgroup, so $S$ is uniquly determined.
In fact $S = \langle c,d \rangle$ with
$$c = (3,5,4,8)(6,7,10,9),\
d = (3,6,4,10)(5,9,8,7).$$
Note that $G$ is $3$-transitive, with no elements fixing more than $2$ points.
Now $N_G(S)$ must have order $16$ and contain an element $e$ outside of $S$
containing the cycle $(1,2)$. Now $e$ must also normalize a subgroup of order
$4$ in $S$, which we will take to be $\langle c \rangle$. (The argument in
the other two cases, $\langle d \rangle$ and $\langle cd \rangle$ is similar.)
By multiplying $e$ by an element of $S$, we may assume that $e$ fixes the point $3$. Since $e$ fixes at most two points, it must invert $\langle c \rangle$,
and hence contains the cycle $(5,8)$.
So there are just two possibilities, for $e$:
$(1,2)(5,8)(6,7)(9,10)$ and $(1,2)(5,8)(6,9)(7,10).$
For the second of these, $be$ fixes $3$ points, which is impossible, so
$$e = (1,2)(5,8)(6,7)(9,10),\ {\rm and}\
G = \langle a,b,c,d,e \rangle.$$
In fact, this really is a group of order $720$, but it is the group $M_{10}$,
which is not simple: the subgroup $\langle a,b,c,e \rangle$ has order $360$.
A proof of the uniqueness of the simple group of order $360$ follows similar
lines to this one, and ends up proving that $G=\langle a,b,c,e \rangle$.
$\mathbf{Proposition}$ (Jordan) If a primitive permutation group $G$ on a set $\Omega$ contains a $3$-cycle then $G$ contains the alternating group ${\rm Alt}(\Omega)$.
Proof. (From Thm 3.3A of "Permutation Groups" by Dixon & Mortimer. It is true even when $\Omega$ is infinite and the proof is not much harder, but I will just prove it in the finite case.) Let $\Delta$ be a largest subset of $\Omega$ such that $G$ contains ${\rm Alt}(\Delta)$ (i.e. the alternating group on $\Delta$ fixing all other points of $\Omega$). By hypothesis we have $|\Delta| \ge 3$. We prove the proposition by showing that $\Delta=\Omega$. Suppose not.
Since $G$ is primitive, there exists $g \in G$ such that $\Delta \cap \Delta^g$ is a non-empty proper subset of $\Delta$. Then $G$ contains ${\rm Alt}(\Delta^g)$. If $\Delta \cap \Delta^g = \{e\}$ has size $1$, then we have $(a,b,e),(c,d,e) \in G$ for any $a,b \in \Delta\setminus \{e\}$, $c,d \in \Delta^g\setminus \{e\}$, and then $(a,e,c) = [(a,b,e),(c,d,e)] \in G$. If $\Delta \cap \Delta^g$ has size bigger than one then we have $(a,b,c) \in G$ with $a,b \in \Delta \cap \Delta^g$ and $c \in \Delta^g \setminus \Delta$.
So in either case, we can find $a,b \in \Delta$ and $c \in \Omega \setminus \Delta$ with $(a,b,c) \in G$. Let $H$ be the subgroup of $G$ generated by ${\rm Alt}(\Delta)$ and $(a,b,c)$. Then $H$ is transitive on $\Delta \cap \{c\}$, and so we have $H={\rm Alt}(\Delta \cap \{c\})$, contradicting the maximality of $\Delta$.
Best Answer
Since $G$ is transitive, $|G|=n\cdot |G_x|$ by the orbit-stabilizer theorem. If $p\mid |G|$ then either $p\mid n$ or $p\mid |G_x|$. Thus $$\pi(G)\leq \pi(n)\cdot \pi(G_x)\leq n\cdot \pi(G_x).$$
That's the first point. For the second, we already have $\pi(G)\leq n\cdot \pi(G_x)$. We also have that $\pi(G_x)\leq (n/2)^{\log(n/2)}$. We are trying to show by induction the following statement:
Thus we can substitute in: $$ \pi(G)\leq n\cdot \pi(G_x)\leq n\cdot (n/2)^{\log(n/2)}.$$ So the question is whether the final term is at most $n^{\log n}$. This isn't really a question about group theory, but it's not hard to see.