Prove that $f(x):=\sqrt{x}$ on $(0,+\infty)$ is $n$-times differentiable.

The answer by @joseville is a classic example of how, when doing induction, it's sometimes easier to prove a stronger statement. So easier than proving that all the derivatives exist, is to prove that they follow a specific formula.

I think that the easiest thing of all is to prove something intermediate: that the $n$th derivative $f^{(n)}(x)$ is a constant multiple of $x^{1/2-n}$. So this is stronger than the statement that the derivatives exist, but weaker than a statement that includes a formula for the constant. (Since the precise constant doesn't affect the proof, this way you don't have to think of the formula for it.)

Also, note that (since you're working directly from the definition) the algebraic manipulations that you do for the first derivative (where the original exponent is positive) will be a little different from the one that you do for all of the subsequent derivatives (where the original exponent is negative).

Here’s how I would do it.

You need an expression for the $n$-th derivative of $f$, so that you can use it to prove that the $(n+1)$th derivative exists (this is the induction (or hereditary) step). So evaluate several derivates and see if you can come up with an expression for the $n$-th derivative.

Let $f^n$ denote the $n$-th derivative of functio $f$ for $n > 0$.

$$ \begin{aligned} f(x) &= \sqrt x = x^{\frac12}\\ f^1(x) &= \frac12 x^{-\frac12}\\ f^2(x) &= -\frac14 x^{-\frac32}\\ f^3(x) &= \frac38 x^{-\frac52}\\ &\;\;\vdots\\ f^n(x) &=\ ??? \end{aligned} $$

Once you have an expression for $f^n(x)$, prove that $f^{n+1}(x)$ exists and go from there.