Find the sum of $\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$ 
Find the sum of $$\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$$


The solution in the book is a lot different than what I tried
what they did in the book is say $k-1=t$ then they expanded from this point $\sum_{t=0}^{n-1} (t+1)(n) {{n-1} \choose {t-1}}$ and the final answer is $n[(n-1)(2^{n-2}-1)+2^n-2]$ but I did not understand their way as it complicated to me
I thought about using derivatives but I got stuck and I am not sure if it really works I just remember using derivatives for some cases in class and I tried it
$(x+b)^n=$ ${n \choose 0} x^0 b^{n-0}+ {n \choose 1} x^1 b^{n-1} +...+ {n \choose n} x^n b^{n-n} $
derivative with respect to $x$ is
$n(x+b)^n-1=$ $0+ {n \choose 1} b^{n-1} +...+ {n \choose n} nx^{n-1} b^{n-n} $
then derivative again
$n(n-1)(x+b)^n-2=$ $0+0+ {n \choose 2} 2b^{n-2} +...+ {n \choose n} n(n-1)x^{n-2} b^{n-n} $
but I could not continue from here
is it ok to use derivatives ? is there another way other than the book(just expanding)?
thanks for any tips and help!

\begin{align}
\sum_{k=1}^n k(k-1) \binom{n}{k-1}
&= \sum_{k=2}^n k(k-1) \binom{n}{k-1} \\
&= \sum_{k=2}^n k n \binom{n-1}{k-2} \\
&= n \sum_{k=2}^n (k-2+2)\binom{n-1}{k-2} \\
&= n \sum_{k=2}^n (k-2)\binom{n-1}{k-2} + 2n \sum_{k=2}^n \binom{n-1}{k-2} \\
&= n (n-1) \sum_{k=3}^n \binom{n-2}{k-3} + 2n \sum_{k=2}^n \binom{n-1}{k-2} \\
&= n (n-1) \sum_{k=0}^{n-3} \binom{n-2}{k} + 2n \sum_{k=0}^{n-2} \binom{n-1}{k} \\
&= n (n-1) \left(2^{n-3} - 1\right) + 2n \left(2^{n-2} - 1\right)\\
&= n (n+3) 2^{n-3} - n (n+1)
\end{align}
