Obtaining the Rodrigues formula On $So(3)$ the algebra of a $3 \times 3$ skew symmetric matrices define Lie bracket $[A,B]=AB-BA$
Consider the exponential map $$EXP: So(3) \to So(3)$$.
We have the $So(3)$ matrix $$A=\begin{bmatrix} 0 & -c & b \\c &  0 & -a\\
-b & a & 0\end{bmatrix}$$
Upon letting $\theta=\sqrt{a^2 + b^2 +c^2}$ , show that we obtain the identity (which is the Rodrigues formula) $$EXP (A)=I_3 + \frac{sin \theta}{\theta} A+ \frac{I-cos \theta}{\theta^2} A^2$$
I am not sure how we get the expressions $$A^{2n}=(-1)^{n+1} \theta^{2(n+1)}\begin{bmatrix} -(b^2+c^2) & ab & ac \\ab &  -(a^2+c^2) & bc\\
ac & bc & -(a^2+b^2)\end{bmatrix}$$ 
and $$A^{2n+1}=(-1)^n \theta^{2n}\begin{bmatrix} 0 & -c & b \\c &  0 & -a\\
-b & a & 0\end{bmatrix}$$
I understand that you look at the powers of $A$, $A^2$, $A^3$ and so on. 
I also understand that you get $A^{2n}$ for even powers of n and $A^{2n+1}$ for odd powers of n. 
I work out $$A^2= \begin{bmatrix} -b^2-c^2 & ab & ac \\ab &  -a^2-c^2 & bc\\
ac & bc & -a^2-b^2\end{bmatrix}$$
$$A^3= \begin{bmatrix} 0 & a^2c-c(-b^2-c^2) & -a^2b+b(-b^2-c^2) 
\\-b^2 c+c(-a^2-c^2) &  0 & ab^2-a(-a^2-c^2)\\
bc^2-b(-a^2-b^2) & -ac^2+a(-a^2-b^2) &  0\end{bmatrix}$$

From $A^2$ how do you get the expression for $A^{2n}$?
From $A^3$ how do you get the expression for $A^{2n+1}$?
For example how is $\theta$ incorporated?

 A: "Rodrigues formula" will be easier to prove with $A/\theta=B$, thus the result we have to prove is under the form: 
$$Exp(\theta B)=I_3 + (\sin \theta) B+ (1-\cos \theta) B^2 \ \ \ (0)$$
by taking this definition:
$$B:=\begin{bmatrix} 0 & -c & b \\c &  0 & -a\\
-b & a & 0\end{bmatrix} \ \ (1a) \ \ \text{with} \ \  \sqrt{a^2+b^2+c^2}=1 \ \ \ (1b)$$
(your presentation has its own merits, but its drawback is that it is un-natural for an angle to have the dimension of a length).
Using (1a) and (1b), the characteristic polynomial of $B$ is found to be $$det(B-\lambda I_3)=-\lambda^3-\lambda \ \ \ $$ 
whose roots (eigenvalues of $B$) are $\{0,i,-i\}$. 
As a consequence, Cayley-Hamilton's theorem gives 
$$B^3=-B \ \ \ \text{and, consequently} \ \ \ B^4=-B^2, B^5=-B^3=B, \ \ (2)  \ \ \text{etc.}$$
(cycling: $-B, -B^2, B, B^2, -B...$). Use now the definition 
$$Exp(\theta B)=\sum_{k=0}^{\infty}\dfrac{1}{k!}\theta^k B^k \ \ \ (3)$$
Transform (3) by using the different relationships (2). Gather them into 3 groups,


*

*the first one, with $I_3$ alone.

*the second group with terms $\alpha_kB$.

*the third group with terms $\beta_kB^2$.
It is easy to see that $\sum_{k=0}^{\infty} \alpha_k=\sin \theta$ and $\sum_{k=0}^{\infty} \beta_k=1-\cos \theta$, proving (0). 
Remark: A similar, simpler, computation exists in 2D. Here it is.  Let $J=\begin{bmatrix}0&-1\\1&0 \end{bmatrix}$ (representing a $\pi/2$ rotation, with eigenvalues $i$ and $-i$). A basic relationship for $J$ is $J^2=-I$, thus $J^3=-J$, etc. Proceeding in the same way as we have done upwards, one obtains 
$$Exp(\theta J)=(\cos\theta)I_2+(\sin\theta)J$$
which is nothing else than a matrix version of the classical $e^{i\theta}=(\cos\theta)+(\sin\theta)i$ but can also be seen and proved as the particular case of (0) with $a=0, b=0, c=1.$
