Show that if $f$ is a polynomial with degree $n$ in two variables, then:
$f(x_0 +x, y_0 +y) = \sum\limits_{i+j \leq n} \frac{1}{i!j!}\frac{\partial^{i+j}}{\partial x^{i} \partial y^{j}} f(x_0,y_0) x^{i} y^{j}$
My attempt: $f$ is a polynomial, then $f(x,y) = \sum\limits_{i+j\leq n}^{n} a_{ij}x^{i}y^{j}$. Then I have to prove that the coefficients have that expression involving partial derivatives. But maybe this problem is out of my level, since I honestly don't see how. Can someone give me a hint? Thanks.
Context: I'm studying for the first time real analysis in $\mathbb{R}^n$. Didn't see yet Inverse Function Theorem.
The idea is good but it would be easier to express $f(x_0 + x, y_0+y)$ as a polynomial in $x$ and $y$. With that in mind, write $\displaystyle f(x_0 + x, y_0+y) = \sum_{0 \le i,j \le n} a_{i,j}x^iy^j$. Then, note that setting $x=y=0$, you get $a_{0,0} = f(x_0,y_0)$. Finally, note that differentiating "kills" the smaller coefficients: $\displaystyle \frac{\partial^{p+q} f}{\partial x^p \partial y^q}(x_0+x,y_0+y) = \sum_{\substack{p \le i \le n\\ q \le j \le n}} \frac{i!j!}{(i-p)!(j-q)!}a_{i,j}x^{i-p}y^{j-q} $. Evaluating this, once again, at $x=y=0$, only the term $a_{p,q}$ remains and we get $\displaystyle \frac{\partial^{p+q} f}{\partial x^p \partial y^q}(x_0,y_0) = p!q!a_{p,q}$.