https://en.wikipedia.org/wiki/Recurrence_relation#Difference_operator_and_difference_equations
I know that $$ \Delta^k a_n = \sum_{t=0}^{k} (-1)^t \binom{k}{t}a_{n+k-t}. $$
But I don't know how to prove that $$ a_{n+k} = a_{n} + \binom{k}{1} \Delta a_n + \cdots + \binom{k}{k} \Delta^{k}(a_n). $$ How to prove it?
On
Here is perhaps a simpler solution. Let $L$ be the left shift operator; given a sequence $a$, $La$ is a new sequence, for which $(La)_n=a_{n+1}$.
Note that $L=\Delta+1$, as operators. This implies the following operator equation: $$ L^k=(\Delta+1)^k=\sum_{i=0}^k \binom ki \Delta^i\tag1 $$ Applying the operators on both sides of $(1)$ to the sequence $a$, and examining the $n^\text{th}$ element, the result is exactly what you want: $$ (L^ka)_n=a_{n+k}=\sum_{i=0}^k \binom ki (\Delta^i a)_n, $$ You can prove by induction on $k$ that $(L^ka)_n=a_{n+k}$; intuitively, the $k$-fold application of the shift-by-one operator is a shift-by-$k$ operator.
For a sequence $a_n$, define the operator $S$ by $$ Sa_n=a_{n+1}\tag1 $$ Then, $$ \begin{align} \Delta^ka_n &=(S-1)^ka_n\tag{2a}\\[9pt] &=\sum_{j=0}^k(-1)^j\binom{k}{j}S^{k-j}a_n\tag{2b}\\ &=\sum_{j=0}^k(-1)^j\binom{k}{j}a_{n+k-j}\tag{2c} \end{align} $$ Explanation:
$\text{(2a):}$ $\Delta=S-1$
$\text{(2b):}$ Binomial Theorem
$\text{(2c):}$ apply $(1)$
Furthermore, $$ \begin{align} \sum_{k=0}^m\binom{m}{k}\Delta^ka_n &=\sum_{k=0}^m\sum_{j=0}^k(-1)^j\binom{m}{k}\binom{k}{j}a_{n+k-j}\tag{3a}\\ &=\sum_{k=0}^m\sum_{j=0}^k(-1)^{k-j}\binom{m}{k}\binom{k}{j}a_{n+j}\tag{3b}\\ &=\sum_{j=0}^m\sum_{k=j}^m(-1)^{k-j}\binom{m}{j}\binom{m-j}{k-j}a_{n+j}\tag{3c}\\ &=\sum_{j=0}^m\binom{m}{j}[j=m]a_{n+j}\tag{3d}\\[6pt] &=a_{n+m}\tag{3e} \end{align} $$ Explanation:
$\text{(3a):}$ apply $(2)$
$\text{(3b):}$ substitute $j\mapsto k-j$
$\text{(3c):}$ $\binom{m}{k}\binom{k}{j}=\binom{m}{j}\binom{m-j}{k-j}$ (expand into ratios of factorials)
$\text{(3d):}$ $\sum\limits_{k=j}^m(-1)^{k-j}\binom{m-j}{k-j}=[j=m]$ (Iverson Brackets)
$\text{(3e):}$ evaluate the sum