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!

Since $(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$, differentiate twice to get
$$n(n-1)(1+x)^{n-2} = \sum_{k=2}^n k(k-1)\binom{n}{k}x^{k-2}$$
Substituting $n+1$ for $n$ we also get
$$(n+1)n(1+x)^{n-1} = \sum_{k=2}^{n+1} k(k-1)\binom{n+1}{k}x^{k-2}.$$
Substitute $x=1$ in each of these, giving
\begin{align*}
n(n-1)2^{n-2} &= \sum_{k=2}^n k(k-1)\binom{n}{k} \\
n(n+1)2^{n-1} &= \sum_{k=2}^{n+1} k(k-1)\binom{n+1}{k}.
\end{align*}
Subtract the first of these from the second, giving
\begin{align*}
\tag{*}2^{n-2}n(2(n+1)-(n-1)) &= \sum_{k=2}^{n+1}k(k-1)\left(\binom{n+1}{k}-\binom{n}{k}\right)\\
&= \sum_{k=2}^{n+1}k(k-1)\binom{n}{k-1}.
\end{align*}
Finally, this gives
$$\sum_{k=1}^{n}k(k-1)\binom{n}{k-1} = n2^{n-2}(n+3) - n(n+1)$$
since the summand of (*) is zero for $k=1$ and is $n(n+1)$ for $k=n+1$. The right-hand side indeed simplifies to
$$n((n-1)(2^{n-2}-1)+2^n-2).$$
