$\begin{aligned}
&\sum_{i=0}^n{n\choose i}p^i(1-p)^{n-i}i^k\\
=&\sum_{i=0}^n{n\choose i}p^i(1-p)^{n-i}\sum_{j=0}^k{i\choose j}j!\begin{Bmatrix}
k \\ j
\end{Bmatrix}\\
=&\sum_{j=0}^k\begin{Bmatrix}
k \\ j
\end{Bmatrix}j!\sum_{i=0}^n{n\choose i}{i\choose j}p^i(1-p)^{n-i}\\
=&\sum_{j=0}^k\begin{Bmatrix}
k \\ j
\end{Bmatrix}j!\sum_{i=0}^n{n\choose j}{n-j\choose i-j}p^i(1-p)^{n-i}\\
=&\sum_{j=0}^k\begin{Bmatrix}
k \\ j
\end{Bmatrix}n^{\underline j}p^j\sum_{i=0}^{n}{n-j\choose i-j}p^{i-j}(1-p)^{n-i}\\
=&\sum_{j=0}^k\begin{Bmatrix}
k \\ j
\end{Bmatrix}n^{\underline j}p^j
\end{aligned}$
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