The trick is to make a little detour by bivariate calculus.
Let $F(x, y)$ be a bivariate function, and recall the chain rule
$$
D_t F(x(t),y(t)) = \dfrac{\text{d} F}{\text{d} a_1} (x,y) D_t x(t) + \dfrac{\text{d}F}{\text{d} a_2}(x, y) D_t y(t)
$$
where $\dfrac{\text{d} F}{\text{d} a_i}(x, y)$ is the derivative of $F$ with respect to its $i$-th variable, evaluated at $x$ and $y$. (I want to avoid any confusion regarding the variables).
This implies that
$$
D_x F(x, x) = \dfrac{\text{d} F}{\text{d} a_1} (x,x)+ \dfrac{\text{d}F}{\text{d} a_2}(x, x). \tag{1}
$$
If I define a linear operator by $\mathcal E_{y\to x} F(x, y) = F(x, x)$, then in operator fashion, (1) becomes
$$
D_x \mathcal E_{y\to x} = \mathcal E_{y\to x} (D_x + D_y), \tag{2}
$$
which is true for all (totally) differentiable functions. By applying $D_x$ on the left and using (2), we get
$$
D_x^2 \mathcal E_{y\to x} = D_x \mathcal E_{y\to x} (D_x + D_y) = \mathcal E_{y\to x} (D_x + D_y)^2,
$$
which can be generalized by induction for integers $n$ to
$$
D_x^n \mathcal E_{y\to x} = \mathcal E_{y\to x} (D_x + D_y)^n. \tag{3}
$$
Finally we can apply this equality (3) to $f(x)g(y)$ to get on the left hand side
$$
D_x^n \mathcal E_{y\to x} f(x) g(y) = D_x^n f(x)g(x),
$$
and on the right hand side
$$
\begin{align}
\mathcal E_{y\to x} (D_x + D_y)^n f(x)g(y) &= \mathcal E_{y\to x} \sum_{k=0}^n \binom{n}{k} D_x^k D_y^{n-k} f(x) g(y) \\
&= \mathcal E_{y\to x} \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(y) \\
&= \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x),
\end{align}
$$
where we have used the binomial theorem, which works because $D_x$ and $D_y$ commute when applied to the space generated by products of the form $l(x)m(y)$ by the linearity of the derivative. Hence the equality
$$
D_x^n f(x)g(x) = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x).
$$
Note: I'd like to mention that this proof is not an "Umbral Calculus" proof, but more of an "Operational Calculus" one, as Umbral Calculus is the study of Sheffer sequences. But both are strongly connected.
The (forward, unit step) finite difference of a function is defined as
$$
\Delta f(x) = f(x + 1) - f(x)
$$
and its iteration as
$$
\Delta ^n f(x) = \Delta \left( {\Delta ^{n - 1} f(x)} \right) = \sum\limits_{0 \le k\left( { \le n} \right)} {\left( { - 1} \right)^{\,n - k} \left( \matrix{
n \hfill \cr k \hfill \cr} \right)f(x + k)}
$$
The binomial
$$
\left( \matrix{ a + x \cr m \cr} \right)
$$
is a polynomial in $x$ of degree $m$,
and the product
$$
\left( \matrix{ a + x \cr m \cr} \right)\left( \matrix{ c + x \cr q \cr} \right)
$$
a polynomial of degree $m+q$.
Your expression is just
$$
\left( { - 1} \right)^{\,k} \Delta ^k
$$
of a polynomial of degree $m_1 + m_2$, which is identically null whenever $k > m_1 + m_2$
Following the suggestion of @jeanmarie and yours, let's explore a bit more the properties of $\Delta$
applied to polynomials.
Let $x^{\,\underline {\,q\,} }$ represents the Falling Factorial.
For $0 \le q \in \mathbb N$ the falling factorial can be defined as
$$
x^{\,\underline {\,q\,} } = x\left( {x - 1} \right)\left( {x - 2} \right) \cdots
\left( {x - q + 1} \right) = \prod\limits_{k = 0}^{q - 1} {\left( {x - k} \right)}
$$
which clearly is a polynomial of degree $q$.
You can easily check that
$$
\eqalign{
& \left| {\;n,q \in N} \right. \cr
& x^{\,\underline {\,0\,} } = 1 \cr
& \Delta x^{\,\underline {\,q\,} } = qx^{\,\underline {\,q - 1\,} } \quad \cr
& \Delta ^n x^{\,\underline {\,q\,} }
= q\left( {q - 1} \right) \cdots \left( {q - n + 1} \right)x^{\,\underline {\,q - n\,} }
= q^{\,\underline {\,n\,} } x^{\,\underline {\,q - n\,} } \quad \left| {\;0 \le n < q} \right. \cr
& \Delta ^q x^{\,\underline {\,q\,} } \equiv q!\quad \quad \left| {\;0 \le n = q} \right. \cr
& \Delta ^n x^{\,\underline {\,q\,} } \equiv 0\quad \left| {\;0 \le q < n} \right. \cr}
$$
i.e. the $\Delta$ operator on $x^{\,\underline {\,q\,} }$ "mimicks" the $D$ operator on $x^q$ ( => umbral calculus, ..)
We can expand the product defining the Falling Factorial as
$$
x^{\,\underline {\,q\,} } = \prod\limits_{k = 0}^{q - 1} {\left( {x - k} \right)}
= \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \, q} \right)}
{\left( { - 1} \right)^{\,q - j} \left[ \matrix{ q \hfill \cr j \hfill \cr} \right]x^{\,j} }
$$
where the square bracket represents the unsigned Stirling N. of 1st kind.
The matrix of the Sitling N. is Lower Triangular and invertible and
$$
x^{\,q} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,q} \right)}
{\left\{ \matrix{ q \hfill \cr k \hfill \cr} \right\}\,x^{\,\underline {\,k\,} } }
$$
where the curly bracket represents the Stirling N. of 2n kind.
So the sequence
$$
\left\{ {x^{\,\underline {\,0\,} } ,x^{\,\underline {\,1\,} } ,\; \cdots ,\,x^{\,\underline {\,q\,} } ,\; \cdots } \right\}
$$
constitutes a polynomial basis (for finite degree at least).
Since $\Delta$ is a linear operator, then
$$
\Delta ^n x^{\,\underline {\,d\,} } \equiv 0\;\quad \Rightarrow \quad
\Delta ^n x^{\,d} \equiv 0\quad \Rightarrow \quad \Delta ^n p_d \left( x \right)
\equiv 0\quad \left| {\;0 \le d < n} \right.
$$
which is the answer given above.
Coming to the Newton series, in my opinion the best way to "grasp it" is through the mutual relation of $\Delta$ with the Shift operator
$$
E:\;\;Ef(x) = f(x + 1)\quad \Leftrightarrow \quad \Delta = \left( {E - I} \right)
$$
so that
$$
\eqalign{
& \Delta ^n = \left( {E - I} \right)^n =
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
{\left( { - 1} \right)^{\,n - k} \left( \matrix{ n \cr k \cr} \right)E^{\,k} } \quad \Leftrightarrow \cr
& \Leftrightarrow \quad \Delta ^n f(x) =
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
{\left( { - 1} \right)^{\,n - k} \left( \matrix{ n \cr k \cr} \right)E^{\,k} f(x)} =
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
{\left( { - 1} \right)^{\,n - k} \left( \matrix{ n \cr k \cr} \right)f(x + k)} \cr}
$$
as we already saw.
We invoke now the Binomial Transform Inversion to write
$$
\eqalign{
& E^n = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
{\left( \matrix{ n \cr k \cr} \right)\Delta ^{\,k} } \quad \cr
& \Leftrightarrow \quad E^n f(x) = f(x + n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)}
{\left( \matrix{ n \cr k \cr} \right)\Delta ^{\,k} f(x)} \cr}
$$
which is the (forward) Newton expansion.
Finally, taking a generic polynomial
$$
p_d (x) = a_{\,d} \,x^{\,d} + a_{\,d - 1} \,x^{\,d - 1} + \; \cdots \; + a_{\,0} \,x^{\,0}
= \sum\limits_{0\, \le \,j\, \le \,d} {a_{\,j} \,x^{\,j} }
$$
we can equivalently write it as
$$
\eqalign{
& p_d (x) = b_{\,d} \,x^{\,\underline d } + b_{\,d - 1} \,x^{\,\underline {d - 1} } + \; \cdots \; + b_{\,0} \,x^{\,\underline 0 }
= \sum\limits_{0\, \le \,j\, \le \,d} {b_{\,j} \,x^{\,\underline {\,j\,} } } = \cr
& = d!b_{\,d} \,\left( \matrix{ x \cr d \cr} \right) + \left( {d - 1} \right)!b_{\,d - 1} \,\left( \matrix{
x \cr d - 1 \cr} \right) + \; \cdots \; + 0!b_{\,0} \left( \matrix{ x \cr 0 \cr} \right) \cr}
$$
Using the relations $x^{\,\underline {\,q\,} } \leftrightarrow \,x^{\,j} $ provided above, and the inversion relation of the Stirling N.
we come to
$$
b_{\,k} = \sum\limits_{0\, \le \,j\, \le \,d} {a_{\,j} \,\left\{ \matrix{ j \hfill \cr
k \hfill \cr} \right\}\;} \quad \Leftrightarrow \quad a_{\,k}
= \sum\limits_{0\, \le \,j\, \le \,d} {\left( { - 1} \right)^{\,j - k} \,\left[ \matrix{
j \hfill \cr k \hfill \cr} \right]b_{\,j} }
$$
which is the premise to the answer signalled by @jeanmarie .
This very succintly is the circle of relations I was mentioning in my comment:
it would have become more apparent if we were to adopt the matrix vectorial representation of polynomials, and base change.
Best Answer
Here we derive a more general identity:
$$ \sum_{j=0}^{m} \binom{m}{j}(x+y)^j \sum_{k=0}^{n} \binom{n}{k} x^k (m-j+n-k)! = \sum_{j=0}^{m} \binom{m}{j} y^{m-j} (j+n)! \sum_{i=0}^{j+n} \frac{x^i}{i!}. \tag{*} $$
The proof is fairly simple and relies on the following identity:
$$ \int_{0}^{\infty} (t+x)^n e^{-t} \, \mathrm{d}t = n!\sum_{i=0}^{n} \frac{x^i}{i!}. $$
The above identity can be proved either by the mathematical induction on $n$ or using the Poisson process. Then
\begin{align*} \text{[LHS of (*)]} &= \sum_{j=0}^{m} \binom{m}{j}(x+y)^j \sum_{k=0}^{n} \binom{n}{k} x^k \int_{0}^{\infty} t^{m-j+n-k}e^{-t} \, \mathrm{d}t \\ &= \int_{0}^{\infty} (t+x+y)^m (t+x)^n e^{-t} \, \mathrm{d}t \\ &= \sum_{j=0}^{n} \binom{m}{j} y^{m-j} \int_{0}^{\infty} (t+x)^{j+n} e^{-t} \, \mathrm{d}t \\ &= \sum_{j=0}^{n} \binom{m}{j} y^{m-j} (j+n)! \sum_{i=0}^{j+n} \frac{x^i}{i!} \\ &= \text{[RHS of (*)]}. \end{align*}