Let $U, V$ be open subsets of $\Bbb R^{n}$ and $\Bbb R^m$ respectively such that $f:U \to V$ is a $C^\infty$ function. If $g:V \to \Bbb R^p$ is also a $C^\infty$ function, how do I show that the composite $g \circ f:U \to \Bbb R^p$ is also a $C^\infty$ function?
My progress: Since $f$ is a $C^\infty$ function, all partial derivatives $\frac{\partial f}{\partial x_j}$ exist on $U$ and are continuous and hence $f$ is differentiable on $U$. Similarly, $g$ is also differentiable on $V$ and thus using the chain rule we find that $g \circ f$ is also differentiable on $U$. Thus, $\frac{\partial (g \circ f)}{\partial x_j} = D_{e_j}(g \circ f)$, the directional derivative along $e_j$ exists.
I can't even see why $\frac{\partial (g \circ f)}{\partial x_j}$ has to be continuous. Now to show that $g \circ f$ is a $C^\infty$ function, we have to show that all partial derivatives $\frac{\partial^k (g \circ f)}{\partial x_{i_1}\partial x_{i_2}\cdots\partial x_{i_k}}$ exist and are continuous. I think I have to use induction but I can't see how I can use it. Any help would be appreciated.
You seem to have some confusion over derivatives in multiple variables. I hope the below clarifies the situation:
Let $f: U \rightarrow V$ be a $C^1$ map, where $U \subseteq \mathbb{R}^n$ and $V \subseteq \mathbb{R}^m$. Then in the standard basis we can view the derivative $Df$ of $f$ as the $m \times n$ Jacobian matrix: $$Df = \begin{bmatrix} \dfrac{\partial f_1}{\partial x_1} & \cdots & \dfrac{\partial f_1}{\partial x_n}\\ \vdots & \ddots & \vdots\\ \dfrac{\partial f_m}{\partial x_1} & \cdots & \dfrac{\partial f_m}{\partial x_n} \end{bmatrix}$$ Since $f$ is $C^1$, all the partial derivatives exist and are continuous. At each point $x \in U$, $Df(x)$ is a linear map from $\mathbb{R}^n$ to $\mathbb{R}^m$.
Similarly, if $g: V \rightarrow \mathbb{R}^p$ is a $C^1$ map, then we can write $$Dg = \begin{bmatrix} \dfrac{\partial g_1}{\partial y_1} & \cdots & \dfrac{\partial g_1}{\partial y_m}\\ \vdots & \ddots & \vdots\\ \dfrac{\partial g_p}{\partial y_1} & \cdots & \dfrac{\partial g_p}{\partial y_m} \end{bmatrix}$$ where again the partial derivatives exist and are continuous.
Then by the chain rule, the derivative $D(g \circ f)$ of the composition $g \circ f$ of $f$ and $g$ is: $$D(g \circ f) = Dg \cdot Df,$$ which at means at each point $x \in U$ that $$D(g \circ f)(x) = Dg(y) \cdot Df(x),$$ where $y = f(x)$. In the standard basis you can view the above as the multiplication of matrices. In summation notation, we can write the partial derivative of $g \circ f$ as $$\frac{\partial (g \circ f)}{\partial x_j} = \sum_{i = 1}^m \left(\frac{\partial g}{\partial y_i}\circ f\right) \cdot \frac{\partial f_i}{\partial x_j}.$$
A moment's thought should convince you that the matrix entries of derivative of $D(g \circ f) = Dg \cdot Df$ in the standard basis exist and are continuous, and therefore $g \circ f$ is $C^1$. In particular, $\partial(g \circ f)/\partial x_j$ is continuous for $j=1,\dots,n$, which at least answers your most basic question.
Now can you finish the argument that if $f$ and $g$ are $C^\infty$, so is $g \circ f$?