About polynomial harmonic functions I am really struggling to solve the following, I don't even know how to start. I would appreciate if anyone could give me some help.

Let $m$ be a positive integer and $u : \mathbb{R}^{n} \rightarrow
 \mathbb{R}$  be a harmonic function. If $u(x) = O(\left|x \right|^m)$
  when $\left|x \right| \to \infty$, show that $u$ is polynomial of
  degree at most $m$.

Thanks in advance for any help.
 A: Use the following analogue of the Cauchy estimates for harmonic functions: if $u:\mathbb{R}^n\to\mathbb{R}$ is harmonic, then for any multi-index $\alpha$ and $r>0$,
\begin{equation}
\vert D^{\alpha}(0)\vert\leq \frac{C_{\alpha}\sup_{\vert x\vert=r} \vert u(x)\vert}{r^{\vert \alpha\vert}},
\end{equation}
where $C_{\alpha}$ is some constant depending only on $\alpha$, and $\vert \alpha\vert$ is the order of $\alpha$. For any multi-index $\alpha$ with $\vert \alpha\vert\geq m+1$, your assumption shows that
\begin{equation}
\vert D^{\alpha}(0)\vert\leq \frac{C'_{\alpha}r^m}{r^{m+1}}\to 0,
\end{equation}
where $C'_{\alpha}$ is some other constant depending only on $\alpha$ as $r\to \infty$, so every partial derivative at $0$ of order at least $m+1$ is zero. For any other $x\in \mathbb{R}^n$, simply shift the function appropriately and use the same argument to show all derivatives of order $\geq m+1$ are zero at $x$.
The reason why the Cauchy estimates hold is by writing $u$ using the Poisson kernel and differentiating under the integral: see page 33 of these notes for a full proof.
