In my textbook I have been given a proving question involving Poisson distribution. The question sentence is as follows:
If $m$ and $\mu_r$ denote the mean and central $r$th moment of a Poisson distribution, then prove that $$\mu_{r+1}=rm\mu_{r-1}+m\cdot\frac{d}{dm}\mu_r$$
What I have tried to do till now is as follows: $$\sum_{x=0}^\infty (x-m)^{r+1}\cdot\frac{e^{-m} m^x}{x!}=rm\sum_{x=0}^\infty (x-m)^{r-1}\cdot\frac{e^{-m} m^x}{x!}+m\cdot\frac{\text{d}}{\text{d}m} \sum_{x=0}^\infty (x-m)^r \cdot \frac{e^{-m} m^x}{x!}$$
Now, since the term $e^{-m}$ is present in each term on L.H.S. and R.H.S. I factor it out and it gets cancelled and I am left with as follows:
$$\sum_{x=0}^\infty \frac{(x-m)^{r+1}\cdot m^x}{x!}=r\sum_{x=0}^\infty \frac{(x-m)^{r-1} m^x}{x!} + m\cdot\frac{\text{d}}{\text{d}m}\sum_{x=0}^\infty \frac{(x-m)^r \cdot m^x}{x!}$$
Solving further gives: $$\sum_{x=0}^\infty \frac{(x-m)^r\cdot(x-m) m^x}{x!} =rm \sum_{x=0}^\infty \frac{(x-m)^{r-1} m^x}{x!} + m\left[-r\sum_{x=0}^\infty \frac{(x-m)^{r-1}\cdot m^x}{x!}+\sum_{x=0}^\infty \frac{(x-m)^r\cdot x\cdot m^{x-1}}{x!}\right]$$
Solving further gives:$$\sum_{x=0}^\infty \frac{(x-m)^r \cdot(x-m) m^x}{x!} = \sum_{x=0}^\infty \frac{(x-m)^r\cdot x\cdot m^x}{x!}$$
Solving further gives:$$\sum_{x=0}^\infty \frac{(x-m)^r \cdot x\cdot m^x}{x!}-\sum_{x=0}^\infty \frac{m \cdot (x-m)^r \cdot m^x}{x!}=\sum_{x=0}^\infty \frac{(x-m)^r\cdot x\cdot m^x}{x!}$$
Solving further gives:$$-\sum_{x=0}^\infty \frac{(x-m)^r\cdot m^{x+1}}{x!} \neq 0$$
I am unable to figure out the mistake I have done. Can you help me in solving this problem from here. Also, is there a chance that the problem has some error?