Why is the composition of smooth multivariable functions smooth? Is an an easy way to see that if $f: U  \subset  \mathbb R^n \rightarrow V  \subset \mathbb R^m$ and $g: V  \subset \mathbb R^m \rightarrow \mathbb R^p$ are smooth functions then their composite $g \circ f: U \rightarrow \mathbb R^p$ is smooth ?
I've been looking at the definiton of the derivative (multivariable) and the chain rule. However, help is needed in order to fully see that $g \circ f$ is smooth.
Looking at the one-dimensional case, I see easily that $g \circ f$ is smooth by elementary rules for differentiation.
 A: If you want hard proof, you're basically ask for a proof of the chain rule. There are many proofs of this and it's not necessary to post one on MSE. Though, if you want help with a particular step in a proof you have you could post that as your question. If intuition is what you ask for, then think of smooth functions as just that, a transformation that neither tears nor creases. If you do one of these, and then another, you still won't be tearing or creasing.
Edit: (In light of the comment)
You need to specify more than "smooth", if you really are asking about higher order derivatives, as it has more than one usage. I'll describe and prove a statement similar to what I think you want, in the case that $f,g$ are both $\mathcal C^{\infty}$. You are right in that the bare chain rule does not state this.
Proposition: If $f, g$ (as you described) are $\mathcal C^{\infty}$ functions, then so is $g\circ f$.
Proof: We need to prove that each partial derivative of $h =g\circ f$ is $\mathcal C^{\infty }$. With respect to standard cartesian coordinates, the chain rule says that $\frac{\partial h_i}{\partial x_j}$ exists for each $1\leq i \leq p$, $1\leq j \leq n$, and is equal to $$\frac{\partial h_i}{\partial x_j} = \sum_{k=1}^m\left(\frac{\partial g_i}{\partial x_k}\circ f\right)\cdot\frac{\partial f_k}{\partial x_j}$$
But all of the partial derivatives of $f$ and of $g$ are $\mathcal C^{\infty}$ by assumption, and we have $\frac{\partial h_i}{\partial x_j}$ expressed as products, sums, and composition with $f$ of these, which we know maintain differentiability. Therefore $h$ is $\mathcal C^{\infty}$.
In the case you only assume $f$ or $g$ are only $\mathcal C^r$, $\mathcal C^s$, respectively, then $h$ will have the least amount of differentiability of the two. For example, suppose $1\leq r\leq s$. Then $h$ will be $\mathcal C^r$, and you can come up with an example where an $(r+1)^{\text{}th}$ partial derivative of $h$ is not continuous (in one dimension this is easy).
