Why this can be happen $1=0$? (With Maclaurin Series)

Question

Suppose I have a function $f(x)=x$

And I can rewrite the function as follows :

$$x=e^{\ln{x}}$$

Rewrite it again in Maclaurin series form:

$$e^{\ln{x}}=\sum_{n=0}^{\infty} \frac{\ln^n{x}}{n!}$$

So we have

$$x=\sum_{n=0}^{\infty} \frac{\ln^n{x}}{n!}$$

When I plug $x=1$ into the equation, then:

$$\begin{align} 1&=\sum_{n=0}^{\infty} \frac{\ln^n{1}}{n!}\\ &=\sum_{n=0}^{\infty} \frac{(\ln{1})^n}{n!}\\ &=\sum_{n=0}^{\infty} \frac{0^n}{n!}\\ &=\sum_{n=0}^{\infty} \frac{0}{n!}\\ &=\sum_{n=0}^{\infty} 0\\ &=0\\ \\ \therefore 1&=0 \end{align}$$

Why this can be happen?

Please tell me. Thanks.

Answer 1

The Maclaurin development of the exponential is

$$\exp(z)=1+z+\frac{z^2}2+\frac{z^3}{3!}+\cdots=1+\sum_{n=1}^\infty\frac{z^n}{n!}$$ so that $\exp(0)=1$.

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Technically, $\dfrac{z^0}{0!}$ is not well defined at $z=0$, safer to write $1$.

Answer 2

Power series such as $\sum_{n=0}^\infty a_nT^n$ work under the assumption that $T^0$ stands for the constant $1$ function. Therefore the RHS is $1$, not $0$.