The exponential maps skew-symmetric matrices onto SO(n)

Inclusion $\exp(\mathrm{Skew}_n)\subset SO(n)$. Let $K^T=-K$. Transposition commutes with the series (it is linear and continuous), so $(e^K)^T=e^{K^T}=e^{-K}=(e^K)^{-1}$, where the last equality uses $e^Ke^{-K}=e^0=I$. Hence $e^K$ is orthogonal. Moreover $\det e^K=e^{\operatorname{tr}K}=e^0=1$ since a skew-symmetric matrix has zero diagonal.

Reverse inclusion. Let $U\in SO(n)$. By the spectral theorem for orthogonal matrices there is $Q\in O(n)$ with
$$U=Q\,\mathrm{diag}\bigl(R_{\theta_1},\dots,R_{\theta_m},\,\varepsilon_1,\dots,\varepsilon_p\bigr)\,Q^T,\qquad\varepsilon_i=\pm1,$$
where the $R_{\theta_j}$ are $2\times2$ rotation blocks (complex eigenvalues $e^{\pm i\theta_j}$ come in conjugate pairs) and the remaining eigenvalues are real, equal to $\pm1$. Since $\det U=1$, the number of $-1$'s among the $\varepsilon_i$ is even. Group them in pairs: $\mathrm{diag}(-1,-1)=R_\pi$. So we may assume that all remaining blocks are rotation blocks $R_{\theta_j}$ (with some $\theta_j=\pi$) and a block of $1$'s:
$$U=Q\,\mathrm{diag}\bigl(R_{\theta_1},\dots,R_{\theta_m},I_{p'}\bigr)\,Q^T.$$
By the computation $e^{\theta J}=R_\theta$ and since the exponential of a block-diagonal matrix is the block-diagonal matrix of the exponentials,
$$\mathrm{diag}(R_{\theta_1},\dots,R_{\theta_m},I)=\exp\bigl(\mathrm{diag}(\theta_1J,\dots,\theta_mJ,0)\bigr).$$
Therefore $U=Qe^{D}Q^T=e^{QDQ^T}$ with $QDQ^T$ skew-symmetric ($D$ is skew-symmetric and $Q$ orthogonal).

Connectedness. $\mathrm{Skew}_n$ is a vector space, hence path connected, and $\exp$ is continuous, so its image $SO(n)$ is path connected.

Remark. The determinant $-1$ component $O(n)\setminus SO(n)$ is not reached by $\exp$ of skew matrices. $O(n)$ has exactly two components, $SO(n)$ and $\det^{-1}(-1)$. In dimension $n=2$, $\exp(\theta J)$ already gives all rotations, and a reflection is not a rotation.