Trying to find $\sum\limits_{k=0}^n k \binom{n}{k}$ 
Possible Duplicate:
How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$? 

$$\begin{align}
&\sum_{k=0}^n k \binom{n}{k} =\\ 
&\sum_{k=0}^n k \frac{n!}{k!(n-k)!} =\\
&\sum_{k=0}^n k \frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!} = \\
&n\sum_{k=0}^n \binom{n-1}{k-1} =\\
&n\sum_{k=0}^{n-1} \binom{n}{k} + n \binom{n-1}{-1} =\\
&n2^{n-1} + n \binom{n-1}{-1}
\end{align}$$


*

*Do I have any mistake?  

*How can I handle the last term?


(Presumptive) Source: Theoretical Exercise 1.12(a), P18, A First Course in Pr, 8th Ed, by S Ross

If you know some elementary calculus, there's a simple way to do this.
By the Binomial theorem we have
$$
(x+1)^n=\sum_{k=0}^n \binom{n}{k}x^k
$$
Differentiate both sides with respect to $x$:
$$
n(x+1)^{n-1}=\sum_{k=0}^n\binom{n}{k}kx^{k-1}
$$
Set $x=1$
$$
n2^{n-1}=\sum_{k=0}^n k\binom{n}{k}
$$
