A neighbourhood of the identity generates the connected group

1. Open subgroup. Let $H=\langle V\rangle$ be the subgroup generated by $V$. Shrink $V$ to an open neighbourhood $V'\subset V$ of the identity. For $h\in H$, the left translate $hV'$ is open (left multiplication by $h$ is a homeomorphism), contains $h$ and is contained in $H$. So $H$ is open.

Closed. The complement of $H$ is the union of the cosets $gH$ with $g\notin H$, each of which is open (translate of the open set $H$). So $H$ is closed.

Conclusion. $H$ is open, closed and nonempty in the connected space $G$, hence $H=G$.

2. Application. $GL_n^+(\mathbb{R})$ is a topological group; it is connected (polar decomposition: $GL_n^+\cong SO(n)\times\mathrm{Sym}_n$). By part 1 it is generated by any neighbourhood of $I$ in $GL_n^+$. The set $\exp(M_n)$ is contained in $GL_n^+$ (since $\det e^A=e^{\operatorname{tr}A}>0$), and by the logarithm near the identity it contains a ball $B(I,\frac12)$. So $\exp(M_n)$ contains a neighbourhood of $I$ in $GL_n^+$ and therefore generates $GL_n^+$.

Concretely: every matrix of positive determinant is a finite product of exponentials: $A=e^{X_1}e^{X_2}\cdots e^{X_m}$ (one needs only $m=2$: polar decomposition gives $A=e^{K}e^{S}$ with $K$ skew, $S$ symmetric).

Remark. The statement is false without connectedness: in $GL_n(\mathbb{R})$ the exponentials generate only $GL_n^+$, never a matrix of negative determinant.