Prove that $\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n + 2)2^{n-1}$. How can I prove the following identity?

$$\binom{n}{0} + 2\binom{n}{1} + ...(n+1)\binom{n}{n} = (n +
 2)2^{n-1}$$

I thought about differentiating this:
$$(1 + x) ^ n = \sum_{k = 0} ^ {n} \binom{n}{k}x^k$$
and then evaluating it at $x = 1$, but I didn't get to my desired result. I just kept finding that $\displaystyle\sum_{k = 0}^n k \binom{n}{k} = n2^{n-1}$.
 A: Here's an alternative proof that does not depend on derivatives:
\begin{align}
\sum_k (k+1) \binom{n}{k} 
&= \sum_k k\binom{n}{k} + \sum_k \binom{n}{k} \\
&= \sum_k n\binom{n-1}{k-1} + \sum_k \binom{n}{k} \\
&= n 2^{n-1} + 2^n \\
&= (n+2) 2^{n-1}
\end{align}
A: You were almost there. Differentiating
$$(1+x)^n=\sum_{k=0}^n{n\choose k}x^k$$
gets you
$$
n(1+x)^{n-1}=\sum_{k=1}^{n}k{n\choose k}x^{k-1}.
$$
Plugging in $x=1$ then yields
$$
\sum_{k=0}^{n}k{n\choose k} = n2^{n-1},
$$
which is as far as you got. However, if we add $\sum_{k=0}^{n}{n\choose k}=2^n=2\cdot 2^{n-1}$ to both sides, the result is
$$
\sum_{k=0}^n(k+1){n\choose k} = (n+2)2^{n-1},
$$
as desired.
A: Another way.
Multiply
$(1+x)^n=\sum_{k=0}^n{n\choose k}x^k
$
by $x$
to get
$x(1+x)^{n}=\sum_{k=0}^n{n\choose k}x^{k+1}
$.
Now when you differentiate
you get
$\begin{array}\\
\sum_{k=0}^n(k+1){n\choose k}x^{k}
&=(x(1+x)^{n})'\\
&=(1+x)^n+xn(1+x)^{n-1}\\
&=(1+x)^{n-1}(1+x+nx))\\
&=(1+x)^{n-1}(1+x(n+1))\\
\end{array}
$
Setting
$x=1$ gives
$\sum_{k=0}^n(k+1){n\choose k}
=(n+2)2^{n-1}
$.
More generally,
Multiply
$(1+x)^n=\sum_{k=0}^n{n\choose k}x^k
$
by $x^m$
to get
$x^m(1+x)^{n}=\sum_{k=0}^n{n\choose k}x^{k+m}
$.
Now when you differentiate
you get
$\begin{array}\\
\sum_{k=0}^n(k+m){n\choose k}x^{k+m-1}
&=(x^m(1+x)^{n})'\\
&=mx^{m-1}(1+x)^n+x^mn(1+x)^{n-1}\\
&=x^{m-1}(1+x)^{n-1}(m(1+x)+xn))\\
&=x^{m-1}(1+x)^{n-1}(m+x(n+m))\\
\end{array}
$
Setting
$x=1$ gives
$\sum_{k=0}^n(k+m){n\choose k}
=2^{n-1}(2m+n)
$.
This, of course,
can be gotten by adding
earlier results.
