Consider functions of the type
$$f(ax) = a^k f(x)$$
where $f: \mathbb{R}^n \to \mathbb{R}$
A simple example is $f(x) = x^k$.
Then I claim,
$$\nabla f(ax)^T(a \delta_i) = a^k \dfrac{\partial f}{\partial x_i} (x)$$
where $\delta_i$ is the vector of all zeroes except for $1$ at the ith position.
Cannot show this. I always have a $a^{k+1}$ on the right hand side. I suspect there might be a typo? Can someone please verify.
There is no typo. For your example you get $\nabla f(x) = k x^{k-1}$, so $\nabla f(ax)a = k(ax)^{k-1}a = a^{k}(kx^{k-1}) = a^k \nabla f(x)$
You apparently made the mistake to go from $$f(ax) = a^k f(x)$$ to$$\nabla f(ax) = a^k \nabla f(x).$$ That is not true, because you think that you take the derivative with respect to $x$ whereas you actually take the derivative w.r.t. $ax$ on the left hand side. To compute $\frac{d}{dx} f(ax)$ you need the chain rule: $f(ax) = f(h(x))$ with $h(x) = ax$, so $\frac{df(ax)}{dx} = \frac{df(h(x))}{dh}\frac{dh}{dx} = \nabla f \cdot a$.