Let $k\in \mathbb N$ and $p>1$. If $v$ solves the ODE
$$\tag{1} \frac{d^kv}{dt^k} = \lvert v\rvert^{p-1}v$$
then $v_\lambda(t):=\lambda^\frac{k}{p-1} v(\lambda t),$ where $\lambda >0$, also solves (1). Let us denote the "scaling exponent" associated with this symmetry by
$$s:=\frac{k}{p-1}.$$
Main observation. For all $T>0$, the function $$\tag{2} v_T(t):=C(T-t)^{-s}$$ solves (1). (The constant $C>0$, not important for this question, is given below).
Is there a conceptual explanation for the appearance of the scaling exponent $s$ in the blow-up solution (2)?
The constant is $$ C= \left((-1)^k(1-s-k)_k\right)^\frac{1}{p-1},$$ where $(n)_k=n(n+1)\ldots (n+k-1).$