Derivation of Kummer's identity I have to derive Kummer’s identity $$e^x\operatorname{M}\left(c-a,c;-x\right)=\operatorname{M}\left(a,c;x\right)$$ from Pfaff’s transformation
$$\operatorname{F}\left(a,b,c;x\right)=\left(1-x\right)^{-b}\operatorname{F}\left(c-a,b,c;\frac{x}{x-1}\right)$$
Attempt: Since, Kummer's function $\operatorname{M}\left(a,c;x\right)$ is the special case of hypergeometric function $\operatorname{F}\left(a,b,c;x\right)$
So, Pfaff’s transformation will be
$$\operatorname{M}\left(a,c;x\right)=\left(1-x\right)\operatorname{M}\left(c-a,c;\frac{x}{x-1}\right)$$
$$=\left(1-x\right)^{-1}\sum_{n=0}^{\infty}{\frac{(c-a)_n}{(c)_nn!}\left(\frac{x}{x-1}\right)^n}\\
=\sum_{n=0}^{\infty}{\frac{\left(1-x\right)^{-1-n}(c-a)_n}{(c)_nn!}\left(-x\right)^n}$$
how does $e^x$ comes in picture
 A: We have
\begin{align*}
M(a,c;x) & = \mathop {\lim }\limits_{b \to  + \infty } F(a,b;c;x/b)
\\ &
 = \mathop {\lim }\limits_{b \to  + \infty } \left( {1 - \frac{x}{b}} \right)^{ - b} F\left( {c - a,b;c;\frac{{x/b}}{{x/b - 1}}} \right)
\\ &
 = \mathop {\lim }\limits_{b \to  + \infty } \left( {1 - \frac{x}{b}} \right)^{ - b} \sum\limits_{n = 0}^\infty  {\frac{{(c - a)_n }}{{(c)_n n!}}\frac{{(b)_n }}{{b^n }}\frac{1}{{(1 - x/b)^n }}\left( { - x} \right)^n } 
\\ &
 = \mathop {\lim }\limits_{b \to  + \infty } \left( {1 - \frac{x}{b}} \right)^{ - b} \sum\limits_{n = 0}^\infty  {\frac{{(c - a)_n }}{{(c)_n n!}}\mathop {\lim }\limits_{b \to  + \infty } \frac{{(b)_n }}{{b^n }}\mathop {\lim }\limits_{b \to  + \infty } \frac{1}{{(1 - x/b)^n }}\left( { - x} \right)^n } 
\\ &
 = e^x \sum\limits_{n = 0}^\infty  {\frac{{(c - a)_n }}{{(c)_n n!}} \cdot 1 \cdot 1 \cdot \left( { - x} \right)^n } 
\\ &
 = e^x M(c - a,c; - x).
\end{align*}
The change in the order of the summation and the limit operation can be justified by Tannery's theorem.
