Prove that $\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n + 2)2^{n-1}$.

You were almost there. Differentiating $$(1+x)^n=\sum_{k=0}^n{n\choose k}x^k$$ gets you $$ n(1+x)^{n-1}=\sum_{k=1}^{n}k{n\choose k}x^{k-1}. $$ Plugging in $x=1$ then yields $$ \sum_{k=0}^{n}k{n\choose k} = n2^{n-1}, $$ which is as far as you got. However, if we add $\sum_{k=0}^{n}{n\choose k}=2^n=2\cdot 2^{n-1}$ to both sides, the result is $$ \sum_{k=0}^n(k+1){n\choose k} = (n+2)2^{n-1}, $$ as desired.

Here's an alternative proof that does not depend on derivatives: \begin{align} \sum_k (k+1) \binom{n}{k} &= \sum_k k\binom{n}{k} + \sum_k \binom{n}{k} \\ &= \sum_k n\binom{n-1}{k-1} + \sum_k \binom{n}{k} \\ &= n 2^{n-1} + 2^n \\ &= (n+2) 2^{n-1} \end{align}