Finding the lie algebra of the symplectic lie group I am having difficulties completing my proof that 
$\text{Lie}(\text{Sp}(2n)) \equiv \mathfrak{sp}(2n) = \{ X \in Gl(2n)\; |\; X^TJ + JX = 0  \}$
Where $J \equiv \begin{bmatrix}0 & \mathbb{1}_n \\ -\mathbb{1}_n & 0\end{bmatrix}$ and $\text{Sp}(2n) \equiv \{ X \in Gl(2n)\; |\; X^TJX = J  \}$
It is simple to show that 
$A \in \mathfrak{sp}(2n) \Rightarrow A \in \{ X \in Gl(2n)\; |\; X^TJ + JX = 0  \}$
holds, since we can differentiate a path $\gamma(t)$ with $\gamma(0)=\mathbb{1}_{n \times n}$ at $t=0$ and we find:
$\begin{align}
\left. \frac{d}{dt} \gamma(t)^T J \gamma(t) \; \right|_{t=0} &= \frac{d}{dt} J = 0
\\&= \left. \dot{\gamma}^TJ\gamma + \gamma^TJ\dot{\gamma} \right|_{t=0} \\
&= \dot{\gamma}^TJ + J\dot{\gamma} = 0
\end{align}$
But now I want to show that 
$A \in \{ X \in Gl(2n)\; |\; X^TJ + JX = 0  \} \Rightarrow A \in \mathfrak{sp}(2n) $
So I look at the smooth paths 
$\gamma_X(t) = \exp(tX)$ in $Sp(2n)$ with $\gamma(0) = \mathbb{1}_{n \times n}$ 
where 
$X\in \{ X \in Gl(2n)\; |\; X^TJ + JX = 0  \}$. 
And I would like to recover the property 
$\exp(tX)^TJ\exp(tX) = J$ for all $t \in [-\epsilon, \epsilon]$
But I am not able to proove it. Do you guys have any ideas how to complete this proof and recover the property of $\text{Sp}$?
I am aware that there are other ways of proving that the lie algebra of the symplectic lie group is given by the above, but I would like to proof it in this manner.
Any help is greatly appreciated :)

Let $X$ be s.t.  $X^T J=-JX$. We want to prove that
$$\exp(tX^T)J=J\exp(-tX), ~\forall t\in\mathbb R$$
where I used that $\exp(tX)^{-1}=\exp(-tX)$.  The case $t=0$ is trivial, so we move to $t\neq 0$.
We expand both sides of the above equation using the definition of the exponential function, arriving at
$$\left(1+tX^T+\cdots+\frac{t^n(X^T)^n}{n!}+\cdots\right)J=J\left(1-tX+\cdots+\frac{(-1)^nt^n X^n}{n!}+\cdots\right).$$
The first non trivial identity is $tX^TJ=-tJX$, which is true by definition of $X$.
At order $n$ we have to prove that
$$\frac{(X^T)^nJ}{n!}=(-1)^n \frac{JX^n}{n!},$$
All we need is associativity of the product of matrices, as
$$\frac{1}{n!}(X^T)^nJ=\frac{1}{n!}(X^T)^{n-1}X^TJ=-\frac{1}{n!}(X^T)^{n-1}JX=\\
-\frac{1}{n!}(X^T)^{n-2}X^TJX=\frac{1}{n!}(X^T)^{n-2}JX^2=\dots=(-1)^n\frac{ JX^n}{n!},$$
as claimed.
