Confirmation of Uncertainty in Indices New Formula? 
I am experimenting relations with regards of the value with uncertainty raised to the $n$th power.
  I came up with this formula: $$(A\pm\alpha)^n=A^n\pm(A^{n-1}n\alpha)$$
  Anyone here able to confirm if correct or not?

I made this equation by myself and when I tested it, the values are similar compared to the method my teacher shows me.
Example: Find $y^5$ given $y=2.0\pm0.1$
Solution using my teacher's method:
$(2.0\pm0.1)^5=(2.0\pm5\%)^5=(32\pm(5\times5\%))=32(\pm25\%)=32(\pm8)$
Solution using the equation above:
$(2.0\pm0.1)^5$ 
$A=2.0$
$\alpha=0.1$ 
$n=5$
Substituting these numbers become: $2.0^5(\pm0.2^{5-1}\times5\times0.1)=32(\pm16\times0.5)=32(\pm8)$
 A: The way we work out this problem is to start by taking a factor of $A^n$ out of the brackets (I'm going to use $A+\alpha$ rather than $A\pm\alpha$ to keep things simple):
$$ (A + \alpha)^n = A^n \left( 1 + \frac{\alpha}{A} \right)^n $$
Now we expand the right hand side using the binomial theorem:
$$ A^n \left( 1 + \frac{\alpha}{A} \right)^n  = A^n \left( 1 + n\frac{\alpha}{A} + \frac{n(n-1)}{2!} \left(\frac{\alpha}{A}\right)^2 + ... \right) $$
Typically the error $\alpha$ is much less than $A$ so the ratio $\alpha/A$ is much less than one. If $\alpha/A$ is much less than one then $(\alpha/A)^2$ is much, much less than one and we can ignore it and all higher powers. Our expression simplifies to:
$$\begin{align}
 (A + \alpha)^n &\approx A^n \left( 1 + n\frac{\alpha}{A} \right) \\
                &= A^n + A^n n\frac{\alpha}{A} \\
                &= A^n + A^{n-1} n\alpha
\end{align}$$
Which is the same expression that you got.
The expression becomes clearer if we use percentage errors. If we write $\varepsilon$ as the percentage error, so e.g. $\varepsilon_A = \alpha/A$, then we get:
$$ \varepsilon_{A^n} = n \varepsilon_A $$
