Q:

Why $\Gamma^\mu_{\beta\delta;\gamma} =\Gamma^\mu_{\beta\delta,\gamma}$?

Gravitation Page 276 Exercise 11.3 solution indicated that 
$$\nabla_\gamma \nabla _\delta e_\beta
=e_{\mu}\Gamma^\mu_{\beta\delta,\gamma} +(e_\nu\Gamma^\nu_{\mu\gamma}) \Gamma^\mu_{\beta\delta}$$
However, using to "Chain rule for gradient" i.e. Eq 10.22 
$$\nabla_\gamma \nabla _\delta e_\beta
=\nabla_{\gamma}(e_{\mu}\Gamma^\mu_{\beta\delta,\gamma})
=e_{\mu} \Gamma^\mu_{\beta\delta;\gamma} +(e_{\mu;\gamma}) \Gamma^\mu_{\beta\delta}$$
where according to equation 10.19, 
$$\Gamma^\mu_{\beta\delta;\gamma} $$ did have indices and have to be corrected. 
Yet the answer somehow indicated that
$$\Gamma^\mu_{\beta\delta;\gamma}=\Gamma^\mu_{\beta\delta,\gamma} $$.
Why $\Gamma^\mu_{\beta\delta;\gamma} =\Gamma^\mu_{\beta\delta,\gamma}$? Was there some identity? How to show that $\Gamma^\mu_{\beta\delta;\gamma} =\Gamma^\mu_{\beta\delta,\gamma}$?

A:

The $\Gamma$'s are not the components of a (1,2)-tensor, so you must be careful here.  If the $\mathbf e$'s are your basis vectors, then $\mathbf e_\beta$ is a vector field $\mathbf X = X^i \mathbf e_i$ with (constant) components $X^i = \delta^i_\beta$.  Accordingly,
$$\nabla_\delta \mathbf e_\beta \equiv \nabla_\delta \mathbf X = \left(\frac{\partial X^\alpha}{\partial x^\delta} + X^\sigma\Gamma^\alpha_{\delta \sigma}\right)\mathbf e_\alpha = \Gamma^\alpha_{\delta \beta} \mathbf e_\alpha$$
For a given choice of $\delta$ and $\beta$, this is a vector field as well.  
It's crucial to understand this, so I'll repeat.  The $\Gamma^\alpha_{\delta\beta}$'s are not the components of a (1,2)-tensor field; as a result, asking about their covariant derivative does not make sense.  However, for a particular choice of $\delta$ and $\beta$, the set of functions $\{\Gamma^0_{\delta\beta},\Gamma^1_{\delta\beta},\Gamma^2_{\delta\beta},\Gamma^3_{\delta\beta}\}$ is the set of components of the vector field $\nabla_\delta \mathbf e_\beta$.
To make things extra clear, let's just let $\mathbf Y = Y^\alpha\mathbf e_\alpha$ where $Y^\alpha = \Gamma^\alpha_{\delta \beta}$ ( which we'll substitute in at the end).  Then we have
$$\nabla_\gamma\nabla_\delta\mathbf e_\beta \equiv \nabla_\gamma \mathbf Y = \left(\frac{\partial Y^\mu}{\partial x^\gamma} + Y^\nu \Gamma^\mu_{\gamma \nu}\right)\mathbf e_\mu = \left(\frac{\partial}{\partial x^\gamma}\Gamma^\mu_{\delta\beta} + \Gamma^\nu_{\delta\beta}\Gamma^\mu_{\gamma\nu}\right)\mathbf e_\mu$$
$$= \left(\Gamma^\mu_{\delta\beta,\gamma}+\Gamma^\nu_{\delta\beta}\Gamma^\mu_{\gamma\nu}\right)\mathbf e_\mu$$

To recap, the "product rule" for covariant derivatives only applies when you're looking at products of components of tensor fields.  The $\Gamma$'s are, in particular, not the components of a $(1,2)$-tensor field, so the notation $\Gamma^{\alpha}_{\ \ \beta\gamma;\delta}$ is not even well-defined.  It is natural for students to see indices and say "Aha! Tensors!" But this instinct must be tempered.
If the $\Gamma^\alpha_{\ \ \delta\beta}$'s were components of a $(1,2)$-tensor field, and if $\mathbf e_\alpha$'s were the components of a $(0,1)$-tensor field, then you would have that
$$\nabla_\gamma\big(\Gamma^\alpha_{\ \ \delta\beta}\mathbf e_\alpha\big)=\Gamma^\alpha_{\ \ \delta \beta;\gamma}\mathbf e_\alpha+\Gamma^\alpha_{\ \ \delta \beta}\mathbf e_{\alpha;\gamma}$$
However, in this case, neither one is true.  $\mathbf e_\alpha$ is a basis vector, and given a specific choice of $\beta$ and $\delta$, the $\Gamma^\alpha_{\delta\beta}$'s  are the components of the vector field $\nabla_\delta \mathbf e_\beta$.  The appropriate thing to do is then to treat them like any other components of a vector field, at which point you get the desired result.

