So I understand how the derivative of $\sin x$ is $\cos x$.
But why can't we use the formula stating the derivative of $x^n = n(x^{n - 1})$ and say that derivative of $\sin x$ ($x$ is $\sin x$ and $n$ is $1$) is $1$?
And why can't we just apply this to everything with a power of $1$ and say the derivative is $1$?
So, a first way of understand that this wouldn't work is to recognize that you could do that with anything and the derivative would be 1. For instance, if we had $10x^2$, well, I could just say, "well, we will treat this all as a unit, and call it $u$. Therefore, I'm just taking the derivative of $u$. Therefore, the derivative would be $1$. In other words, following your procedure, every derivative would wind up being 1.
So, to use the method you are probably learning (I use a different one, but that's okay), the derivative of $x^n$ is not 1, but rather, it depends on what variable you are taking the derivative with respect to. If $q$ is the variable we are taking the derivative with respect to, the result is actually $nx^{n-1}\frac{dx}{dq}$. That fraction at the end is actually always there, with the numerator being the variable in your formula, and the denominator being the variable you are taking the derivative with respect to. However, in the case of taking the derivative with respect to $x$, this will wind up being $\frac{dx}{dx}$, which is $1$, which means that it can essentially be ignored.
So, going back to your problem. First of all, you can't say that $x = \sin(x)$ because you used the same variable for two different things. You need to choose a new variable. We will choose $u$. This is important, because otherwise everything gets really confusing. So, if we say that $u = \sin(x)$, then, if we take the derivative of $u$ with respect to $x$, we get $1\cdot u^0 \frac{du}{dx} = 1\cdot\frac{du}{dx}$. This is certainly a true value, but it has not gotten us any closer to finding the derivative.
In any case, you will wind up needing a separate "derivative form" for each major function/operation that you use. Otherwise, your results will wind up with weird derivatives like $\frac{du}{dx}$ that are unhelpful.
Good job thinking about substitutions, though! That is definitely the first place a mathematician should go when thinking through problems. The key is to recognize when they lead to good places vs. problematic results.
Your error seems to be expecting the rule$$\frac{d}{dx}\sin x^n=\sin\frac{d}{dx}x^n.$$But this isn't right. We say $\frac{d}{dx}$ doesn't commute with $\sin$. Indeed, it doesn't commute with functions in general. To use an example you already understand,$$\frac{d}{dx}(x^n)^k=nkx^{nk-1}\not\equiv(nx^{n-1})^k=n^kx^{nk-k}.$$
What does work, by something called the chain rule, is$$\frac{d}{dx}\sin x^n=nx^{n-1}\sin^\prime x^n.$$This leaves the derivative $\sin^\prime$ unidentified. (It also tells us nothing we didn't already know about the special case $n=1$.) It turns out $\sin^\prime y=\cos y$, so$$\frac{d}{dx}\sin x^n=nx^{n-1}\cos x^n.$$
To me it seems that OP wanted to apply the power derivative rule $(x^n)' = n x^{n-1}$ for $n \geq 1$ to the following:
$$(\sin x)^1$$
to wrongly obtain $1(\sin x)^0 = 1$. (The current other answers assume OP meant $\sin(x^1)$.)
However, the power derivative rule is not applicable as $\sin x$ and $x$ are syntactically different expressions, you the rule does not even pattern-match. Still, if we slightly rewrite $(\sin x)^1$, we can apply it indirectly.
Define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = x^1$. Then $$(\sin x)^1 = (f \circ \sin)(x)$$ for all $x \in \mathbb{R}$. We can now apply the chain rule to get
$$(f \circ \sin)^{'}(x) = (f' \circ \sin)(x) \cdot \sin'(x) = f'(\sin(x)) \cdot \cos(x) = 1 \cdot \cos(x)$$
Now note that we rediscover what we wrongly used above, namely $f' \equiv 1$. The crucial difference, however, is that we account for $\sin'$ as well here by fully formally applying the chain rule.
To conclude: Never forget the "inner derivative" of the chain rule.
Funnily enough, in German we call that "nachdifferenzieren" ("to post-differentiate"). I haven't encountered an English term yet.
