Exponential of Pauli Matrices 
Let $\vec{v}$ be any real three-dimensional unit vector and $\theta$ a real number. Prove that $\exp(i\theta \vec{v}\cdot\vec{\sigma}) = \cos(\theta)I + i\sin(\theta)\vec{v}\cdot\vec{\sigma}$, where $\vec{v}\cdot\vec{\sigma} \equiv \sum_{k=1}^{3}v_{k}\sigma_{k}$.

$\vec{v}\cdot\vec{\sigma}$ is a scalar so let this scalar be a non-zero integer $n$. 
Effectively, this gives $\exp(i\theta n)$. 
$\exp(i\theta n) \Rightarrow (\cos(\theta) + i \sin(\theta))^{n}$ 
I would appreciates if anyone could provide to me a hint to this.
The computation becomes messy very quickly.
Thanks in advance.
 A: Here's an option

Commutation rules

I'm not aware of a fast way of proving this, other than actually calculate all the products directly, but in any case it is fairly straightforward to show that ([ab]using Einstein's notation)
\begin{eqnarray}
[\sigma_a, \sigma_b] &=& 2i\epsilon_{abc}\sigma_c \\
\{\sigma_a, \sigma_b \} &=&  2 \delta_{ab}I \tag{1}
\end{eqnarray}
From here note that
\begin{eqnarray}
[\sigma_a,\sigma_b] +\{\sigma_a, \sigma_b \}  &=& (\sigma_a\sigma_b - \sigma_b\sigma_a) + (\sigma_a \sigma_b + \sigma_a \sigma_b) \\
2i\epsilon_{abc}\sigma_c + 2 \delta_{ab}I &=& 2\sigma_a \sigma_b \\
\sigma_a \sigma_b &=& \delta_{ab}I + i\epsilon_{abc}\sigma_c \tag{2}
\end{eqnarray}

Vector product

Now consider two vectors ${\bf u}$ and ${\bf v}$ and multiply both sides of Eqn. (2) by their components, contracting indices leads to
\begin{eqnarray}
(u_a v_b) \sigma_a \sigma_b &=& (u_a v_b) \delta_{ab} I + i(u_a v_b) \epsilon_{abc} \sigma_c \\
(u_a \sigma_a) (v_b \sigma_b) &=& u_a v_a I + i (\epsilon_{abc} u_a v_b) \sigma_c \\
({\bf u}\cdot \mathbf{\sigma}) ({\bf v}\cdot \mathbf{\sigma}) &=& ({\bf u}\cdot {\bf v})I + i ({\bf u}\times {\bf v})_c \sigma_c \\
({\bf u}\cdot \mathbf{\sigma}) ({\bf v}\cdot \mathbf{\sigma})&=& ({\bf u}\cdot {\bf v})I + i ({\bf u}\times {\bf v}) \cdot \sigma \tag{3}
\end{eqnarray}
Set ${\bf u} = {\bf v} = \hat{\bf n}$ in the last equation, you will get
$$
({\bf u}\cdot \mathbf{\sigma})^2 = I \tag{4}
$$
And from this it is trivial to see
\begin{eqnarray}
({\bf u}\cdot \mathbf{\sigma})^{2k} &=& I \\
({\bf u}\cdot \mathbf{\sigma})^{2k + 1} &=& ({\bf u}\cdot \mathbf{\sigma}) ~~~\mbox{for}~~~ k = 0,1,\cdots \tag{5}
\end{eqnarray}

Putting everything together

\begin{eqnarray}
e^{i\theta \hat{\bf u}\cdot \sigma} &=& \sum_{k = 0}^{+\infty} \frac{(i\theta)^k}{k!}(\hat{\bf u}\cdot \sigma)^k \\
&=& \sum_{k = 0}^{+\infty} \frac{i^{2k}\theta^{2k}}{(2k)!}(\hat{\bf u}\cdot \sigma)^{2k} + \sum_{k = 0}^{+\infty} \frac{i^{2k+1}\theta^{2k+1}}{(2k+1)!}(\hat{\bf u}\cdot \sigma)^{2k+1} \\
&\stackrel{(5)}{=}& \sum_{k = 0}^{+\infty} \frac{(-1)^k\theta^{2k}}{(2k)!}I + \sum_{k = 0}^{+\infty} \frac{i(-1)^k\theta^{2k + 1}}{(2k + 1)!}(\hat{\bf u}\cdot \sigma) \\
&=& \cos \theta I + i(\hat{\bf u}\cdot \sigma) \sin \theta
\end{eqnarray}
A: The title hints at a crucial bit of missing information: the definition of the Pauli matrices, $\vec\sigma$.  The most common representation is
\begin{align}
\sigma_1 &= \left(\begin{array}{rr}
0&1\\1&0
\end{array}\right)
&
\sigma_2 &= \left(\begin{array}{rr}
0&i\\-i&0
\end{array}\right)
&
\sigma_3 &= \left(\begin{array}{rr}
1&0\\0&-1
\end{array}\right)
&
\end{align}
but the important parts of the definition are the cyclic product $\sigma_1 \sigma_2 = i\sigma_3$ (and permutations) and $\sigma_i\sigma_i = I$.  These are equivalent to the commutation relationships in caverac's answer.
So your statement that 

$\vec v\cdot \vec\sigma$ is a scalar

isn't correct.  In this representation, $\vec v\cdot\vec\sigma$ would be the $2\times2$ matrix, 
$$
\vec v\cdot\vec\sigma = \left(\begin{array}{cc}
v_3 & v_1 + i v_2
\\
v_1 - i v_2 & -v_3
\end{array}\right)
$$
What does it mean to expenentiate a matrix?  Well, the exponential for real numbers is equivalent to a power series:
$$
\exp x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_n \frac{x^n}{n!}
$$
That power series is a set of instructions that we can totally apply to a matrix.  Let's see what happens if we try it with $\sigma_1$:
\begin{align}
\exp i\theta\sigma_1
&= I + i\theta\sigma_1 - \frac{\theta^2 I}{2!} - i\frac{\theta^3 \sigma_1}{3!} + \cdots
\\ &= I \times \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} + \cdots
\right)
+
i\sigma_1 \times \left(
\theta - \frac{\theta^3}{3!} + \cdots
\right)
\\&= I\cos\theta + i\sigma_1\sin\theta
= \left(\begin{array}{cc}
\cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta
\end{array}\right)
\end{align}
This shows you the shape of the result to expect as you complete the proof for general $\vec v\cdot\vec\sigma$.
