Laplace-Beltrami operator in $\mathbb{R}^m$ The Laplace-Beltrami operator, $\Delta\colon \Omega^{k}(M)\longrightarrow \Omega^{k}(M)$ is defined as $\Delta=d\delta + \delta d$ where $d$ is the usual exterior derivative and $\delta$ is the codifferential: $$\delta=(-1)^{m(k+1)+1}*d*$$ where $*$ is the Hodge operator, $\alpha \wedge *\beta= \langle\alpha,\beta\rangle \text{Vol}$.
I am trying to prove that if $f\in C^{\infty}(M)$, $\Delta f=-\sum_{i=1}^{m}\frac{\partial^2f}{\partial x_{i}^2}$.
I have tried to do this: $\Delta f=(\delta d)(f)$ and  $$(\delta d)(f)=\delta\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}dx_{i}.$$
Now, $*(\frac{\partial f}{\partial x_{i}}dx_{i})=(-1)^{i-1}\frac{\partial f}{\partial x_{i}}dx_{1}\wedge \cdots dx_{i-1}\wedge dx_{i+1}\wedge \cdots \wedge dx_{m}.$
Again, 
$$\begin{split}
d \left((-1)^{i-1}\frac{\partial f}{\partial x_{i}}dx_{1}\wedge \cdots dx_{i-1}\wedge \cdots \wedge dx_{m}\right)&=(-1)^{i-1}\frac{\partial ^2 f}{\partial x_{i}^2}dx_{i}\wedge dx_{1}\wedge \cdots \wedge dx_{m}\\
&=\frac{\partial ^2 f}{\partial x_{i}^2}dx_{1}\wedge \cdots \wedge dx_{m}
\end{split},$$ and 
$$*\left(\frac{\partial ^2 f}{\partial x_{i}^2}dx_{1}\wedge \cdots \wedge dx_{m}\right)=\frac{\partial ^2 f}{\partial x_{i}^2}.$$
In conclusion, I obtain that $\Delta f=(-1)^{m+1}\sum_{i=1}^{m}\frac{\partial ^2 f}{\partial x_{i}^2}$, instead of $-\sum_{i=1}^{m}\frac{\partial ^2 f}{\partial x_{i}^2}$.
Can someone help me finding the wrong step? Thanks in advance.

Note the ambiguity: when you write $\delta$, it's really $\delta :\Omega^k \to \Omega^{k-1}$. In the case for $\Delta = \delta d$, $\delta$ is adding on one forms and thus $k=1$. So 
$$ \delta = (-1)^{m(k+1)+1} * d* = (-1)^{2m+1} *d*= - *d*.$$ 
