I got stuck with the following equation
$ze^{zx} = \sum_{j=1}^{\infty} x^{j-1} \frac{z^j}{(j-1)!}$
Have got to this
$z e^{zx} = \sum_{n=0}^{\infty} \frac{z(zx)^n}{n!}$
How can I show that the $n=0$ term is zero in the last equation ?
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It's not; it's $z$. – J.G. Aug 09 '23 at 22:13
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Your approach seems okay to me! I don’t think the $n=0$ term is zero? It would be $z$ which equals zero when $z=0$. – JayP Aug 09 '23 at 22:16
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Ok. Have kept the $n=0$ because as you say, the first term is $z$ rather than $0$. After substituting $n+1=j$, I get the proof. – Veak Aug 09 '23 at 22:30