understanding adjoint representation

Let $G$ be a matrix Lie group like say $\textrm{GL}_n(\Bbb{C})$ or $\textrm{SL}_n(\Bbb{C})$. An alternative way to define the (real) Lie algebra $\mathfrak{g}$ is to say that it is the set of all $n\times n$ complex matrices $X$ such that

$$e^{tX} \in G \hspace{4mm} \textrm{for all $t \in \Bbb{R}$}.$$

Now using the Lie product formula and some other stuff one can check that the set of all such matrices is in fact a linear space, and we make it into a Lie algebra by defining the commutator of two elements $X,Y$ as $[X,Y] = XY - YX$ where by $XY$ I mean the product of two matrices. This is the reason why I chose $G$ to be a matrix Lie group in the first place so that the bracket is something concrete.

Now I am not so familiar with differential geometry but I will try to explain why $\mathfrak{g}$ now is the tangent space at the identity. Now since everything is happening inside of $\Bbb{C}^{n^2}$ we can think of the tangent space at the identity concretely as the set of all $n \times n$ complex matrices $X$ ( vectors in $\Bbb{C}^{n^2}$ ) such that there is a smooth curve $\gamma(t) \in G$ with $\gamma (0) = I$ and $\gamma'(0) = X$.

We can now prove that given the definition of $\mathfrak{g}$ in terms of the exponential, it is indeed equivalent to being the tangent space at the identity. On one hand if $X \in \mathfrak{g}$ then we can simply take the smooth curve

$$\gamma(t) = e^{tX}$$

that is in $G$ for all $t$ and is such that $\gamma(0) = I$, $\gamma'(0) = X$. For the other direction, suppose that there is a smooth curve $\rho(t)$ in $G$ such that $\rho(0) = I$ and $\rho'(0) = Y$, where $Y$ is some matrix. We wish to show that $Y\in \mathfrak{g}$. Now because the exponential map is a local diffeomorphism about $0 \in\mathfrak{g}$ and $I \in G$,, for $t$ sufficiently small we have that

$$\log( \rho(t)) = \rho(t) - I - \frac{(\rho(t) - I)^2}{2} + \frac{(\rho(t) - I)^3}{3} + \ldots $$

is in $\mathfrak{g}$. Now absolute convergence in the Hilber-Schmidt norm allows us to differentiate log term by term and $\mathfrak{g}$ being a linear space means that the derivative at $t = 0$ is in $\mathfrak{g}$. However the derivative at $t = 0$ is nothing more than the matrix $Y$, so consequently $Y \in \mathfrak{g}$.

This completes the proof that the original definition of the Lie algebra that I gave you is equivalent to it being the tangent space at the identity.