Finding the lie algebra of the symplectic lie group

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.