This is a try to fill the holes in this previous exercise.
Let $f$ defined as $f(x):=(x-a)^n g(x)$, for some $g$ continuous in a neighborhood of $a$, we can say that $f$ is $n$ times differentiable at $a$?
If we define the quotient difference operator
$$\Delta^n_h f(x):=\frac1{h^n}\sum_{k=0}^n\binom{n}{k}f(x+(n-k)h)(-1)^k$$
then if $f$ is continuous in a neighborhood of $a$ and if Im not wrong we can write
$$\lim_{h\to 0}\Delta^n_h f(a)=f^{(n)}(a)$$
For the case of the title this reduces to
$$\begin{align}f^{(n)}(a)&=\lim_{h\to 0}\frac1{h^n}\sum_{k=0}^n\binom{n}{k}(-1)^k((n-k)h)^ng(a+(n-k)h)\\&=\lim_{h\to 0}\sum_{k=0}^n\binom{n}{k}(-1)^k(n-k)^ng(a+(n-k)h)\\&=\sum_{k=0}^n\binom{n}{k}(-1)^k(n-k)^ng(a)\end{align}$$
and for derivatives for $j\le n$ we have that $f^{(j)}(a)=0$.
It is this reasoning correct?
The reasoning is not correct. The equality $$\lim\limits_{h\rightarrow 0}\Delta_h^nf(a)=f^{(n)}(a)$$ is true, provided that the RHS exists. However, just because the limit on the LHS exist, it doesn't follow the RHS exists. Indeed, the statement in the question is false. This is because $f(x)$ can oscillate very rapidly in the neighbourhood of $a$, which would make the first derivative be discontinuous at $a$. To be precise, consider for example $$g(x)=x\sin x^{-2}\text{ for }x\neq 0,\\g(0)=0$$ (check it's continuous) and $$f(x)=x^2g(x)=x^3\sin x^{-2}.$$
We then find $$f'(x)=3x^2\sin x^{-2}-2\cos x^{-2},\\ f'(0)=0,$$ which is discontinuous at zero, hence $f$ is not twice differentiable.
For general $n$, you can similarly take $f(x)=x^{n+1}\sin x^{-n}$.