Whether you consider it a deep, very deep or shallow result:

> Theorem. The symplectic group ${\bf Sp}_{2n}({\mathbb R})$ is included in ${\bf SL}_{2n}({\mathbb R})$.

One proof is purely algebraic and uses the Pfaffian (as a matter of fact, the same result is true when one replaces ${\mathbb R}$ by another field).

The other proof is more familiar. Using the polar decomposition, one proves that ${\bf Sp}_{2n}({\mathbb R})$ is diffeomorphic to ${\mathbb R}^\ell\times{\bf U}_n$. Because the unitary group is connected, one obtains the connectedness of the symplectic group. We conclude with the fact that the determinant can take only the dicrete values $\pm1$.

**Edit**. Here is the Pfaffian proof. Recall that the Pfaffian is a polynomial $Pf$ in the entries of the $2n\times2n$ alternate matrix, such that
$$Pf(A)^2=\det A,\qquad Pf(J_n)=1,\quad J_n:=\begin{pmatrix} 0_n & I_n \\\\ -I_n & 0_n \end{pmatrix}.$$
It has the fundamental property that if $P\in{\bf M}_{2n}$, then
$$Pf(M^TAM)=Pf(A)\cdot\det M.$$
Apply this identity to $A=J_n$ and to $M$ a symplectic matrix, you obtain $1=1\cdot\det M$. This proof is in my book *Matrices* (Springer GTM **216**).