[proofplan]
We first compute the effect of one finite difference on a single monomial: it lowers degree by at least one, and if the increment is nonzero then the new leading coefficient is multiplied by the old degree and by the increment. We then iterate this calculation on the leading term $\alpha_k n^k$. The remaining terms of $P$ start in degree at most $k-1$, so after $r$ differences they have degree at most $k-r-1$ and cannot change the coefficient of $n^{k-r}$.
[/proofplan]
[step:Compute the leading term after one finite difference of a monomial]
Let $j\in\mathbb N\cup\{0\}$, let $c\in K$, and define the monomial polynomial $M_{c,j}\in K[n]$ by
\begin{align*}
M_{c,j}(n):=c n^j.
\end{align*}
If $j=0$, then $\delta_h M_{c,0}=0$ for every $h\in K$.
Assume now that $j\ge 1$. Expanding $(n+h)^j$ in $K[n]$ gives
\begin{align*}
(\delta_h M_{c,j})(n)=c((n+h)^j-n^j)=c\sum_{\ell=0}^{j-1}\binom{j}{\ell}h^{j-\ell}n^\ell.
\end{align*}
Hence $\delta_h M_{c,j}$ has degree at most $j-1$. Its coefficient of $n^{j-1}$ is $cjh$, because the term $\ell=j-1$ contributes $c\binom{j}{j-1}hn^{j-1}=cjhn^{j-1}$. If $c\ne 0$ and $h\ne 0$, then $cjh\ne 0$: here $j1_K\ne 0$ because $K$ has characteristic $0$. Therefore $\delta_h M_{c,j}$ has degree exactly $j-1$ and leading coefficient $cjh$.
[guided]
We isolate the one-step calculation because every later step is just this calculation repeated. Fix $j\in\mathbb N\cup\{0\}$ and $c\in K$, and define the polynomial $M_{c,j}\in K[n]$ by
\begin{align*}
M_{c,j}(n):=c n^j.
\end{align*}
If $j=0$, then $M_{c,0}$ is constant, so
\begin{align*}
(\delta_h M_{c,0})(n)=M_{c,0}(n+h)-M_{c,0}(n)=c-c=0.
\end{align*}
Thus a finite difference of a constant polynomial is the zero polynomial.
Now suppose $j\ge 1$. In the [polynomial ring](/page/Polynomial%20Ring) $K[n]$, the binomial expansion gives
\begin{align*}
(n+h)^j=\sum_{\ell=0}^{j}\binom{j}{\ell}h^{j-\ell}n^\ell.
\end{align*}
Subtracting the $\ell=j$ term, which is exactly $n^j$, leaves
\begin{align*}
(\delta_h M_{c,j})(n)=c((n+h)^j-n^j)=c\sum_{\ell=0}^{j-1}\binom{j}{\ell}h^{j-\ell}n^\ell.
\end{align*}
Every power of $n$ appearing in this sum has exponent at most $j-1$, so $\delta_h M_{c,j}$ has degree at most $j-1$.
The coefficient of the highest possible power $n^{j-1}$ comes from the index $\ell=j-1$. That coefficient is
\begin{align*}
c\binom{j}{j-1}h=cjh.
\end{align*}
If $c\ne 0$ and $h\ne 0$, this coefficient is nonzero. The only extra point to check is that $j$ is nonzero as an element of $K$; this is precisely where the characteristic $0$ hypothesis is used. Therefore, under $c\ne 0$ and $h\ne 0$, the degree is exactly $j-1$ and the leading coefficient is $cjh$.
[/guided]
[/step]
[step:Iterate the degree drop for arbitrary polynomials]
For an integer $d\ge 0$, let $K[n]_{\le d}$ denote the set of polynomials in $K[n]$ of degree at most $d$, together with the zero polynomial. Also set $K[n]_{\le d}:=\{0\}$ for $d<0$.
The one-step computation implies that
\begin{align*}
\delta_h(K[n]_{\le d})\subset K[n]_{\le d-1}
\end{align*}
for every $h\in K$ and every integer $d\ge 0$, because $\delta_h$ is $K$-linear and each monomial of degree at most $d$ is sent to a polynomial of degree at most one less.
Applying this containment successively gives
\begin{align*}
\delta_{h_1,\dots,h_r}(K[n]_{\le d})\subset K[n]_{\le d-r}
\end{align*}
for every $r\in\mathbb N$ and every integer $d\ge 0$. Since $P\in K[n]_{\le k}$, it follows that
\begin{align*}
\delta_{h_1,\dots,h_r}P\in K[n]_{\le k-r}.
\end{align*}
Thus $\delta_{h_1,\dots,h_r}P$ has degree at most $k-r$.
[/step]
[step:Track the contribution of the original leading term]
Define $R\in K[n]$ by
\begin{align*}
R(n):=P(n)-\alpha_k n^k.
\end{align*}
Then $R\in K[n]_{\le k-1}$. By linearity of each $\delta_h$,
\begin{align*}
\delta_{h_1,\dots,h_r}P=\delta_{h_1,\dots,h_r}(\alpha_k n^k)+\delta_{h_1,\dots,h_r}R.
\end{align*}
The preceding degree-drop result gives
\begin{align*}
\delta_{h_1,\dots,h_r}R\in K[n]_{\le k-1-r}.
\end{align*}
Hence $R$ contributes no term of degree $k-r$.
It remains to compute the degree $k-r$ coefficient of $\delta_{h_1,\dots,h_r}(\alpha_k n^k)$. Applying the one-step leading-term calculation successively gives leading coefficients
\begin{align*}
\alpha_k k h_1,
\end{align*}
then
\begin{align*}
\alpha_k k(k-1)h_1h_2,
\end{align*}
and after $r$ steps
\begin{align*}
\alpha_k k(k-1)\cdots(k-r+1)h_1\cdots h_r.
\end{align*}
At the $i$-th step, the current leading degree is $k-i+1$, so the finite difference with increment $h_i$ multiplies the current leading coefficient by $(k-i+1)h_i$.
[/step]
[step:Use nonzero increments and characteristic zero to get exact degree]
Assume $h_1\cdots h_r\ne 0$. Then each $h_i\ne 0$. Since $K$ is a field of characteristic $0$, each integer $k-i+1$ with $1\le i\le r$ is nonzero in $K$. Therefore
\begin{align*}
\alpha_k k(k-1)\cdots(k-r+1)h_1\cdots h_r\ne 0.
\end{align*}
The coefficient of $n^{k-r}$ in $\delta_{h_1,\dots,h_r}P$ is exactly this nonzero element, and the lower-degree part $\delta_{h_1,\dots,h_r}R$ cannot cancel it because it has degree at most $k-r-1$. Hence $\delta_{h_1,\dots,h_r}P$ has degree exactly $k-r$ with leading coefficient
\begin{align*}
\alpha_k k(k-1)\cdots(k-r+1)h_1\cdots h_r.
\end{align*}
This proves both assertions.
[/step]
