Sum of combinations formula Is there an explicit formula for the sum $0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+n\dbinom{n}{n} = \sum_{k=0}^nk\dbinom{n}{k}$?

Actually, I think I got it because I realized I forgot to show what work I have already done in the question but as I was trying to refine my work I realized what I need to do:
$$0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+n\dbinom{n}{n}=$$
$$1\dbinom{n}{1}+2\dbinom{n}{2}+\dots+n\dbinom{n}{n}=$$
$$1\cdot\dfrac{n!}{1!\cdot (n-1)!}+2\cdot\dfrac{n!}{2!\cdot (n-2)!}+\dots+n\cdot\dfrac{n!}{n!\cdot 0!}=$$
$$n\left(1\cdot\dfrac{(n-1)!}{1!\cdot (n-1)!}+2\cdot\dfrac{(n-1)!}{2!\cdot (n-2)!}+\dots+n\cdot\dfrac{(n-1)!}{n!\cdot 0!}\right)=$$
$$n\left(\dfrac{(n-1)!}{0!\cdot (n-1)!}+\dfrac{(n-1)!}{1!\cdot (n-2)!}+\dots+\dfrac{(n-1)!}{(n-1)!\cdot 0!}\right)=$$
$$n\left(\dbinom{n-1}{0}+\dbinom{n-1}{1}+\dots+\dbinom{n-1}{n-1}\right)=$$
$$n\left(2^{n-1}\right)=\boxed{n \cdot 2^{n-1}}$$

Based on Lucas Henrique and Lucas De Souza's answers, this is what I came up with:
$$f(x)=(x+1)^n=\dbinom{n}{0}x^n+\dbinom{n}{1}x^{n-1}+\dots+\dbinom{n}{n}x^0$$
Taking the derivative of $f(x)$, we get 
$$f'(x)=n\cdot(x+1)^{n-1}=n\dbinom{n}{0}x^{n-1}+(n-1)\dbinom{n}{1}x^{n-2}+\dots+0\dbinom{n}{n}$$
This can also be written as
$$f'(x)=n\dbinom{n}{n}x^{n-1}+(n-1)\dbinom{n}{n-1}x^{n-2}+\dots+0\dbinom{n}{0}$$
because $\binom{n}{k}=\binom{n}{n-k}$. Plugging in $1$ for $x$, we get
$$f'(1)=n\dbinom{n}{n}+(n-1)\dbinom{n}{n-1}+\dots+0\dbinom{n}{0}$$
as desired. Thus, our sum can be represented by $f'(1)$, which can be calculated as $n\cdot(1+1)^{n-1}=\boxed{n\cdot2^{n-1}}$.
