How will matrix $A^n$ affect the original eigenvector and eigenvalue? For example, a matrix $A$ has three distinct eigenvalues and has 3 eigenvectors $v_1,v_2,v_3$ corresponding to the three distinct eigenvalues. 
So, am I right to say $A^4$ will result in eigenvalue^4 and the eigenvectors remain unchanged?

The claim that the eigenvectors remain unchanged can be interpreted in two different ways, one of which is true, while the other is false.
The first interpretation is

If $v$ is an eigenvector of $A$, then $v$ is an eigenvector of $A^4$.

This claim is true, and easily verified:
If $v$ is an eigenvector of $A$, then there exists a value $\lambda$ such that $Av=\lambda v$. But then $A^4v = A^3Av = A^3(\lambda v) = \lambda A^3v = A^2 A v = \ldots = \lambda^4 v$. Therefore $v$ is also an eigenvector of $A^4$, with the corresponding eigenvalue $\lambda^4$.
The second interpretation is:

$v$ is an eigenvalue of $A^4$ iff $v$ is an eigenvalue of $A$.

That claim is false. A simple counterexample is $A=\pmatrix{1&0\\0&-1}$. Up to a scalar factor, the only eigenvectors of this matrix are $\pmatrix{1\\0}$ and $\pmatrix{0\\1}$. But $A^4$ is the identity matrix, and all non-zero vectors are eigenvectors of the identity.
With the eigenvalues it is even more complicated, as here the claim may depend on whether you consider real or complex vector spaces/matrices.
For example, consider the matrix $A=\pmatrix{2&0&0\\0&0&-1\\0&1&0}$. In the real numbers, the only eigenvalue is $2$. But $A^4$ has eigenvalues $2^4=16$ and $1$.
On the other hand, in the complex numbers, this matrix has the additional eigenvalues $i$ and $-i$, whose fourth powers are, indeed, both $1$.
