A helpful proof in contracting the Christoffel symbol? 
Out of all of my time learning General relativity, this is the one identity that I cannot get around.
  $$ \Gamma_{\alpha \beta}^{\alpha} = \partial_{\beta}\ln\sqrt{-g} \tag{1}$$
  where $g$ is the determinant of the metric tensor $g_{\alpha \beta}$.

With the Christoffel symbol, we start by contracting
$$ \begin{align}
 \Gamma_{\alpha \beta}^{\alpha} &= \frac{1}{2} g^{\alpha\gamma} (\partial_{\alpha} g_{\beta\gamma}  + \partial_{\beta} g_{\alpha\gamma}  - \partial_{\gamma} g_{\alpha\beta} ) \\
&= \frac{1}{2} g^{\alpha\alpha} ( \partial_{\beta} g_{\alpha\alpha}) \\
&= \frac{1}{2g_{\alpha\alpha}} ( \partial_{\beta} g_{\alpha\alpha}) 
\end{align}\tag{2}$$ 
where I took $\gamma \rightarrow \alpha$ and $g^{\alpha\alpha} = 1/g_{\alpha\alpha}$.
The next steps to take now, I have no clue. MTW gives a hint by saying to use the results from some exercise, which are,
$$\det A = \det||A^{\lambda}_{\ \ \rho}|| = \tilde{\epsilon}^{\alpha\beta\gamma\delta}A^{0}_{\ \ \alpha}A^{1}_{\ \ \beta}A^{2}_{\ \ \gamma}A^{3}_{\ \ \delta} $$
$$(A^{-1})^{\mu}_{\ \ \alpha}(\det A) = \frac{1}{3!}\delta_{\alpha\beta\gamma\delta}^{\mu\nu\rho\sigma} A^{\beta}_{\ \ \nu} A^{\gamma}_{\ \ \rho}A^{\delta}_{\ \ \sigma} $$
$$ \mathbf{d}\ln|\det A| = \mathrm{trace}(A^{-1}\mathbf{d}A) ,\tag{3}$$
where $\mathbf{d}A$ is the matrix $||\mathbf{d}A^{\alpha}_{\ \ \mu}||$ whose entries are one-forms.
I fail to reason why the metric turns into the determinant from what I have done and then becomes the result at the top.
 A: The original and the most general definition of determinant is given by Gauss
. For the determinant of metric tensor we write
\begin{eqnarray}
 g&:=& \frac{1}{4!}\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon^{{\mu\nu}{\rho\sigma}}g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}g_{\delta\sigma}.\\
 \therefore \delta g &=& \frac{1}{3!}\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon^{{\mu\nu}{\rho\sigma}}g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}\delta g_{\delta\sigma},\\
 &=&\frac{-g}{3!}\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon^{{\mu\nu}{\rho\sigma}} g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}\delta g_{\delta\sigma},\\
 &=&\frac{-g}{3!}\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}\gamma}{}^\sigma \delta g_{\delta\sigma},\\
 &=&g \, g^{\delta\sigma}\delta g_{\delta\sigma}.
\end{eqnarray}
(Equivalently, $$ \delta(\ln |\det g|)=Tr (g^{-1}\delta g)= Tr ( \delta \ln g )$$)
By using this result we the have
$$ \frac{1}{g} \partial_\beta g =g^{\delta\sigma}\partial_\beta g_{\delta\sigma} $$
Note: You may need some basic manipulation of these quantities, $\varepsilon$ is Levi-Civita symbol, $\epsilon$ is Levi-Civita tensor.
$$\epsilon^{{\alpha\beta}{\gamma\delta}} = -\frac{1}{\sqrt{-g}}\varepsilon^{{\alpha\beta}{\gamma\delta}}$$
$$\epsilon_{{\alpha\beta}{\gamma\delta}} = \sqrt{-g}\varepsilon_{{\alpha\beta}{\gamma\delta}}$$
$$\varepsilon_{{\alpha\beta}{\gamma\delta}}=\varepsilon^{{\alpha\beta}{\gamma\delta}}= \delta^{[\alpha}_0 \delta^\beta_1 \delta^\gamma_2 \delta^{\delta]}_3$$
$$\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}} =-\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\mu\nu}{\rho\sigma}}= 4! \delta^{[\alpha}_\mu \delta^\beta_\nu \delta^\gamma_\rho \delta^{\delta]}_\sigma \equiv \delta^{\alpha\beta\gamma\delta}_{\mu\nu\rho\sigma}\equiv \left| \begin{matrix}
\delta^\alpha_\mu  & \delta^\alpha_\nu  & \delta^\alpha_\rho  & \delta^\alpha_\sigma  \\
\delta^\beta_\mu  & \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
\delta^\gamma_\mu  & \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
\delta^\delta_\mu  & \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma 
\end{matrix} \right|$$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\nu}{\rho\sigma}} = -\left| \begin{matrix}
\delta^\alpha_\alpha  & \delta^\alpha_\nu  & \delta^\alpha_\rho  & \delta^\alpha_\sigma  \\
\delta^\beta_\alpha  & \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
\delta^\gamma_\alpha  & \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
\delta^\delta_\alpha  & \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma 
\end{matrix} \right| =-\left| \begin{matrix}
1  & 0  & 0  & 0  \\
0  & \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
0  & \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
0  & \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma 
\end{matrix} \right| =-\left| \begin{matrix}
 \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
 \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
 \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma 
\end{matrix} \right|$$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\rho\sigma}} = -\left| \begin{matrix}
 \delta^\beta_\beta  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
 \delta^\gamma_\beta  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
 \delta^\delta_\beta  & \delta^\delta_\rho  & \delta^\delta_\sigma 
\end{matrix} \right|=-\left| \begin{matrix}
 2  & 0  & 0  \\
 0  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
 0  & \delta^\delta_\rho  & \delta^\delta_\sigma 
\end{matrix} \right|=-2 \left| \begin{matrix}
 \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
  \delta^\delta_\rho  & \delta^\delta_\sigma 
\end{matrix} \right| $$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\gamma\sigma}} =-2 \left| \begin{matrix}
 \delta^\gamma_\gamma  & \delta^\gamma_\sigma  \\
  \delta^\delta_\gamma  & \delta^\delta_\sigma 
\end{matrix} \right| =-2 \left| \begin{matrix}
 3  & 0  \\
  0  & \delta^\delta_\sigma 
\end{matrix} \right| = -3! \delta^\delta_\sigma  $$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\gamma\delta}} =-3! \delta^\delta_\delta = -4!$$
Note2:
\begin{eqnarray}
 g^{-1}&:=& \frac{1}{4!}\varepsilon_{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}}g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}g^{\delta\sigma}.\\
 \therefore \delta g^{-1} &=& \frac{1}{3!}\varepsilon_{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}}g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}\delta g^{\delta\sigma},\\
 &=&\frac{-g^{-1}}{3!}\epsilon_{{\alpha\beta}{\gamma\delta}}\epsilon_{{\mu\nu}{\rho\sigma}} g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}\delta g^{\delta\sigma},\\
 &=&\frac{-g^{-1}}{3!}\epsilon_{{\alpha\beta}{\gamma\delta}}\epsilon^{{\alpha\beta}\gamma}{}_\sigma \delta g^{\delta\sigma},\\
 &=&g^{-1} \, g_{\delta\sigma}\delta g^{\delta\sigma}.\\
 \therefore -g^{-2}\delta g &=&g^{-1} \, g_{\delta\sigma}\delta g^{\delta\sigma}.\\
 \delta g &=& -g\,g_{\delta\sigma}\delta g^{\delta\sigma}. 
\end{eqnarray}
By comparing this $\delta g$ to the first result, we have
$$\delta g_{\alpha\beta}=- g_{\alpha\mu}g_{\beta\nu} \delta g^{\mu\nu}$$
A: Recall the matrix identity 
$$\tag{1}\log\det M=\operatorname{tr}\log M.$$
If $M=M(\lambda)$ is differentiable in $\lambda$, then 
$$\tag{2}\frac{d}{d\lambda}\log\det M=\operatorname{tr}\left(M^{-1}\frac{d}{d\lambda} M\right).$$
The proof of $(1)$ for symmetric matrices follows from the usual formulae for the trace and determinant in terms of eigenvalues$^{1}$.
As for the Christoffels, we have 
$$\Gamma^i{}_{ij}=\frac{1}{2}g^{ik}(\partial_i g_{jk}+\partial_j g_{ik}-\partial_k g_{ij})=\frac{1}{2}g^{ik}\partial_j g_{ik}=\frac{1}{2}\operatorname{tr}(g^{-1} \partial_j g).$$
The last equality is just what the contraction of indices means for the (symmetric!) matrix $g=(g_{ij})$, and there is an error in the indices in OP's post. Now, using $(2)$ we have 
$$\Gamma^i{}_{ij}=\frac{1}{2}\partial_j\log \det g.$$
This can be brought into the form
$$\Gamma^i{}_{ij}=\partial_j \log\sqrt{|\det g|}$$
by the usual rules of calculus.

$^{1}$ For symmetric matrices, such as $g$, it is easy because $g$ can be diagonalized. For other matrices you might need a Jordan normal form to compute $\log M$.
