Binomial summation $\sum\binom nkk^2$

Here's one way to do it:

Recall $$(1 + x)^n = \sum_{k = 0}^n \binom{n}{k}x^k.$$

Plugging in $x = 1$ gives that the sum of the binomial coefficients is $2^n$. Notice that $$x \frac{d}{dx} \left(x \frac{d}{dx} \left( (1 + x)^n\right) \right) = \sum_{k = 0}^n k^2 \binom{n}{k} x^k .$$

Computing the left-hand side and evaluating at $x = 1$ will give you the answer.

There are $n$ people. You want to select some of them to form a committee. Amongst the committee members, you choose a person to be the president. You also pick from the committee members one person (not necessarily different from the president) to be the treasury. Thus, if the committee consists of $k$ members, then the number of ways to pick the president and the treasury is $k^2$. That is, the total number of ways to pick the committee and the two extra positions is $$\sum_{k=0}^n\,k^2\,\binom{n}{k}\,.$$

On the other hand, you can pick the president and the treasury first. In the case that the president and the treasury are different, there are $n$ ways to choose the president, $n-1$ ways to pick the treasury, and $2^{n-2}$ ways to select the remaining committee members, whence there are $n(n-1)\,2^{n-2}$ ways in total. If the president is the same as the treasury, then there are $n$ ways to choose this person for two positions, and $2^{n-1}$ ways to complete the committee, making $n\,2^{n-1}$ ways in total.

Combining the two ways of counting, we conclude that $$\sum_{k=0}^n\,k^2\,\binom{n}{k}=n(n-1)\,2^{n-2}+n\,2^{n-1}=n(n+1)\,2^{n-2}\,.$$ The claim is now proven.

From your procedure:

$$\sum_{k=0}^n\binom nkk^2=n\sum_{k=1}^n\binom{n-1}{k-1}{(k\color{red}{-1+1})} = n\left( (n-1)\sum_{k=2}^{n} \binom{n-2}{k-2}+\sum_{k=1}^{n} \binom{n-1}{k-1}\right)$$

The sum is thus similar to

$$n\left( (n-1)\sum_{k=0}^{n-2} \binom{n-2}{k}+\sum_{k=0}^{n-1} \binom{n-1}{k}\right) = {n((n-1)2^{n-2}+2^{n-1})}= \color{blue}{n(n+1)2^{n-2}}$$