Prove using mathematical induction $1\cdot2+2\cdot2^2+3\cdot2^3+\ldots+n\cdot2^n=2[1+(n-1)2^n]$ 
Prove the result using Mathematical Induction $$1\cdot2+2\cdot2^2+3\cdot2^3+\ldots\ldots+n\cdot2^n=2[1+(n-1)2^n].$$

I've been stuck on this problem for hours, I have no idea how do even calculate it. The exponents throw me off. If anyone can help me break it down step-by-step, I would truly appreciate it.
Here's my attempt


Hint:
The base case is fairly easy: $0\times\left(\ldots\right)=2\left(1+(0-1)\times2^0\right)$. If we assume that $\sum_{n=0}^k n2^{n}=2\left(1+(k-1)\cdot2^{k}\right)$, then our goal for the induction case is to prove that: $$
\underbrace{2\left(1+(k-1)\cdot2^{k}\right)}_{\sum_{n=0}^k {n2^{n}}}+ \underbrace{(k+1)2^{k+1}}_{\text{$(k+1)$'th term}}=2\left(1+k\cdot2^{k+1}\right)$$
Then, divide through by 2 and expand. Can you take it from here?
