I will present a sketch of a proof for the following by induction:
$$\int_{-\infty}^{\infty}\delta(x)^{n}f(x)\mathrm{d}x = {(-1)^n}f^{(n-1)}(0); n>1$$
I will also use the definition
$$\delta(x):=\Theta'(x)$$
Where
$$\Theta(x) = \begin{cases}1 \text{ if }x\geq0\\0\text{ otherwise}\end{cases}$$
I'm not really sure if the steps are right or even if the above is correct, so I need help verifying if this is valid and what kind of function $f$ could be.
Case n=1
$$\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x = f(0) = (-1)^0f^{(0)}(0)$$
Case n+1
$$\int_{-\infty}^{\infty}\delta^{n+1}(x)f(x)\mathrm{d}x = \int_{-\infty}^{\infty}\left(\Theta'(x)\right)^{n+1}(x)f(x)\mathrm{d}x = \left(\Theta'(x)\right)^nf(x)\Bigg{|}_{-\infty}^{\infty} - \int_{-\infty}^{\infty}\left(\Theta'(x)\right)^{n}f^{(1)}(x)\mathrm{d}x$$
By definition of the delta function, $\Theta'(x)=0$ for $x\neq0$, so
$$\int_{-\infty}^{\infty}\delta^{n+1}(x)f(x)\mathrm{d}x = - \int_{-\infty}^{\infty}\left(\Theta'(x)\right)^{n}f^{(1)}(x)\mathrm{d}x = - \int_{-\infty}^{\infty}\delta(x)^nf^{(1)}(x)\mathrm{d}x$$
Using the hypothesis
$$-\int_{-\infty}^{\infty}\delta(x)^nf^{(1)}(x)\mathrm{d}x = -(-1)^{n}f^{(n)}(0) = (-1)^{n+1}f^{n}(0)$$
$\blacksquare$
This is incorrect. One can not work with powers of the Dirac delta function, as if it were a ordinary function. To see that such a calculation goes wrong, let us use the simplest representation of the delta function, namely a block. I.e. we define the delta function $\delta(x)$ as $1/\epsilon$ for $-\epsilon /2 < x < +\epsilon /2$ and $0$ elsewhere. We now get:
$$\int _{-\infty} ^{\infty} \delta(x)^n f(x)dx = \frac {1}{\epsilon ^n} \int_{-\epsilon/2} ^{\epsilon/2}f(x)dx = f(0) / \epsilon^{n-1}$$
As required we now take the limit of $\epsilon$ to $0$, and then we see that for all powers of $n$ larger than $1$ the result diverges, i.e. goes to $+\infty$.