Sum of the rows of Pascal's Triangle. I've discovered that the sum of each row in Pascal's triangle is $2^n$, where $n$ number of rows. I'm interested why this is so. Rewriting the triangle in terms of C would give us 
              $0C0$ in first row.
$1C0$ and $1C1$ in the second, and so on and so forth. However, I still cannot grasp why summing, say, 4C0+4C1+4C2+4c3+4C4=2^4. 

This can also be proven by induction. In order to do so, it is very helpful to prove the following identity:
$$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$$
This can be proven combinatorially, but for the sake of expediency,
\begin{align*}
\binom{n}{k} + \binom{n}{k+1} &= \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!} \\
&= \frac{n!}{k!(n-k)(n-k-1)!} + \frac{n!}{(k+1)k!(n-k-1)!} \\
&= \frac{n!}{k!(n-k-1)!}\left(\frac{1}{n-k} + \frac{1}{k+1}\right) \\
&= \frac{n!}{k!(n-k-1)!}\left(\frac{k+1}{(n-k)(k+1)} + \frac{n - k}{(n-k)(k+1)}\right) \\
&= \frac{n!}{k!(n-k-1)!} \cdot \frac{n+1}{(n-k)(k+1)} \\
&= \frac{(n+1)n!}{(k+1)k!(n-k)(n-k-1)!} \\
&= \frac{(n+1)!}{(k+1)!(n-k)!} \\
&= \binom{n+1}{k+1}.
\end{align*}
Now, let's launch into the induction proof. First, note that $\binom{0}{0} = 1 = 2^0$, so the base case holds.
Next, assume that $\binom{n}{0} + \ldots + \binom{n}{n} = 2^n$. Then, using the identity,
\begin{align*}
&\binom{n+1}{0} + \binom{n+1}{1} + \binom{n+1}{2} + \ldots + \binom{n+1}{n} + \binom{n+1}{n+1} \\
= \, &\binom{n+1}{0} + \left(\binom{n}{0} + \binom{n}{1} \right) + \left(\binom{n}{1} + \binom{n}{2} \right) + \ldots + \left(\binom{n}{n-1} + \binom{n}{n} \right) + \binom{n+1}{n+1}.
\end{align*}
Note that each term of the form $\binom{n}{k}$ where $1 \le k \le n-1$ occurs exactly twice in the above sum. Also, the other terms, $\binom{n+1}{0}, \binom{n}{0}, \binom{n+1}{n+1}, \binom{n}{n}$ are all $1$. Thus we have,
\begin{align*}
&\binom{n+1}{0} + \binom{n+1}{1} + \binom{n+1}{2} + \ldots + \binom{n+1}{n} + \binom{n+1}{n+1} \\
= \, &2\left(\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n-1} + \binom{n}{n} \right) = 2 \cdot 2^n = 2^{n+1}.
\end{align*}
