Conjecture: $(x^\alpha+p)^\beta=\,_1F_0(\beta;;x^\alpha+p-1)$ and ideas for proof

Question

Conjecture: For $|z|<1,\,\alpha,\beta,p\in\Bbb R$ $$(z^\alpha+p)^\beta=\,_1F_0(\beta;;z^\alpha+p-1)$$ I found this formula by noting that $$z^\alpha=\,_1F_0(\alpha;;z-1)$$ Via the simplification of $$z^\alpha=\sum_{n=0}^{\infty}(z-1)^n\frac{\prod_{k=1}^{n}(\alpha-k+1)}{n!}$$ Which is the Taylor series of $z^\alpha$. Hence, I set $$u=z^\alpha+p$$ Thus, I conjectured, $$u^\beta=\,_1F_0(\beta;;u-1)=\,_1F_0(\beta;;z^\alpha+p-1)$$ Is such a maneuver valid? In other words, is it 'legal' to do this substitution in this context?

Answer 1

${_1\hspace{-2px}F_0}$ indeed reduces to a binomial expansion: $${_1\hspace{-2px}F_0}(\beta; ; x) = \sum_{k = 0}^\infty (\beta)_k \frac {x^k} {k!} = \sum_{k = 0}^\infty \binom {\beta + k - 1} k x^k = \sum_{k = 0}^\infty (-1)^k \binom {-\beta} k x^k = (1 - x)^{-\beta}.$$ $x$ stands for any quantity, the only requirement is that the absolute value of that quantity is less than one.