Use mathematical induction to prove that the sum of the entries of the $k^{th}$ row of Pascal’s Triangle is $2^k$. 
Use mathematical induction to prove that the sum of the entries of the $k$-th row of Pascal’s Triangle is $2^k$. Begin by proving that the row sum for any particular row is double that for the previous row.

I am having a hard time trying to figure out how to prove that the row sum for any particular row is double that for the previous row. 
I know how to show for row one, row two and so forth but once I get to row n I know that the sum has to be row(n-1)(2), but I have no idea how to prove that. I know that each row's sum can be written as $2^k$ where $k$ is the row number. I was wondering if anyone can give me a hint or start me off.

There problem go from $n$ to $n+1$:
$$S_{n+1}={n+1\choose 0}+{n+1\choose 1}+...+{n+1\choose n+1}$$
Using Stiefel's rule:
$${n+1\choose k}={n\choose k}+{n\choose k-1}$$ 
and note that 
$${n+1\choose 0}={n\choose 0}, \quad {n+1\choose n+1}={n\choose n}$$
so,
$$S_{n+1}={n\choose 0}+\left[{n\choose 0}+{n\choose 1}\right]+\left[{n\choose 1}+{n\choose 2}\right]+...+\left[{n\choose n-1}+{n\choose n}\right]+{n\choose n}=\\
=2\cdot \left[{n\choose 0}+{n\choose 1}+...+{n\choose n}\right]=2\cdot 2^n=2^{n+1}$$
