Question: The year 2016 is coming up, so it makes me wonder: how do you prove that there is no simple group of order 2016? (without invoking the 10,000 page theorem on the classification of simple groups)

Answer: 
Let $|G| = 2016 = 2^{5}\cdot 3^{2} \cdot 7$.

By Sylow theorems let $n\_{7}$ be the number of Sylow $7$-subgroups. 

Then by Sylow theorems, if $n\_{7} \neq 1$.

We proceed through cases:

Case 1: Suppose $n\_{7} = 8$.

We have $[G: N\_{G}(P)] = n\_{7} = 8$, where $P$ is a sylow $7$-subgroup.
Then $\left|N\_{G}(P)\right| = 4\cdot 9 \cdot 7$.

Now, let $G$ act on the set of left cosets of $N\_{G}(P)$, then this action affords a homomorphism: $\phi: G \rightarrow S\_{8}$.

Now, suppose $G$ is simple, then $\phi$ is injection into $S\_{8}$. If $G$ has a subgroup of index $2$, then it would not be simple and so $G$ is a subgroup of $S\_{8}$ which does not have a subgroup of index $2$ it follows from the Second Isomorphism Theorem that $G$ must be in $A\_{8}$. 

Fact 1: if $P$ is a Sylow $7$-subgroup in $S\_{8}$, then $\left|N\_{A\_{8}}(P)\right| = \frac{1}{2}\left|N\_{S\_{8}}(P) \right|$. 

Fact 2: $\left|N\_{S\_{p +1}}(P) \right| = p(p-1)$, it follows that $|N\_{S\_{8}}(P)| = 6 \cdot 7$. 

It follows from Fact 1 that $\left|N\_{A\_{8}}(P) \right| = 21$. But $N\_{G}(P) \leq N\_{A\_{8}}(P)$ and so we should have $\left |N\_{G}(P) \right|$ dividing $21$, which is absurd and so $G$ must not be simple.
I just stumbled on this question, so I thought I'd "close the loop", and take care of the one case not yet handled in the other answer or the comments: when $n\_7$ is $36$.

Let $P=\langle g\rangle$ be a Sylow 7-group, and suppose $G$ is a simple group with $36$ Sylow 7-subgroups. Then $|N\_G(P)|=56$ and by the normalizer-centralizer theorem, $|C\_G(P)|=28$.

If $z\in C\_G(P)$ is an involution (has order $2$), then $gz$ has order $14$. If we look at the permutation representation $\rho:G\rightarrow A\_{36}$, induced by conjugation of the Sylow 7-subgroups, then we must have (up to conjugation)
$$ \rho(gz) = (1,2,3,4,5,6,7)(8,9,\cdots,21)(22,23,\cdots,35) $$

This is because we need $\rho(gz)$ to be an even permutation, and we need its square $\rho(g)$ to fix exactly one point [the one corresponding to $P$].

Thus $\rho(z)=\rho(gz)^7$ has $8$ fixed points, and thus $z$ normalizes $8$ Sylow 7-subgroups. Note that an element $h\in N\_G(P)\setminus C\_G(P)$ could only fix at most $6$ points, since if it fixed two points in one orbit of $\rho(g)$ then $\rho([h,g])\in\rho(P)$ would fix more than one point, and we would then have $[h,g]=1$, so $h\in C\_G(P)$.

So we really have $z$ *centralizes* $8$ Sylow 7-subgroups. Consider then $C\_G(z)$. It contains $C\_G(P)$, and we must have $[C\_G(z):C\_G(P)]$ divisible by $8$, since $C\_G(P)$ will normalize $P$ in $C\_G(z)$, and there are $8$ Sylow 7-subgroups in $C\_G(z)$. Thus $[G:C\_G(z)]$ divides $9$, and by simplicity of $G$ we get that $[G:C\_G(z)]=9$.

Now $\sigma:G\rightarrow A\_9$ is then a permutation representation, given by conjugation of $z$. Since $\sigma(g)$ is then a 7-cycle, we see that $g$ centralizes exactly $2$ conjugates of $z$, say $z$ and $z^x$. But since being conjugate is an equivalence relation and $z$ was an arbitrary involution in $C\_G(P)$, we see that the involutions of $C\_G(P)$ can be grouped in pairs. This is the contradiction, since an even-order group always has an odd number of involutions.