We have $$e^{2\pi i n}=1$$

So we have $$e^{2\pi in+1}=e$$

which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$ Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$

This implies $$e^{-4\pi^{2}n^{2}}=1$$

Taking the limit when $n\rightarrow \infty$ gives $0=1$.

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    $\begingroup$ I took the liberty of doing an edit that actually changed the meaning of your question, in the middle formula. It made no sense as it stood, but it does now, I think. $\endgroup$ – Harald Hanche-Olsen Aug 4 '12 at 11:55
  • $\begingroup$ well, the reason for the joke is this does not make sense. $\endgroup$ – Bombyx mori Aug 4 '12 at 12:01
  • $\begingroup$ Do you want to undo my edit? You certainly may do so, if you insist. $\endgroup$ – Harald Hanche-Olsen Aug 4 '12 at 12:02
  • $\begingroup$ You actually show that 1 is equal to infinitely many other numbers (if the steps of the argument are correct) $\endgroup$ – i. m. soloveichik Aug 4 '12 at 12:08
  • $\begingroup$ well, we are here to please everyone. $\endgroup$ – Bombyx mori Aug 4 '12 at 12:09

Your error is (as in most of those fake-proofs) in the step where you use the power law $(a^b)^c=a^{bc}$ without the conditions of that power law being fulfilled.

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    $\begingroup$ It could be useful to recall these conditions, or give a reference. $\endgroup$ – enzotib Aug 4 '12 at 13:12
  • $\begingroup$ But such conditions are not easy to produce, except for some special cases: For example, having $a$ and $b$ real (with $a$ nonnegative) is sufficient, unless you base your powers on a highly unusual branch of the natural log function. $\endgroup$ – Harald Hanche-Olsen Aug 4 '12 at 14:15

The proof is wrong because an expression of the form $x^y$ is actually ambiguous, when $x$ is a complex number: Rewrite it as $e^{y\ln x}$ and note the multivalued nature of the natural logarithm as used on complex numbers. For your proof to be correct, you would need $\ln e^{2\pi in+1}=2\pi in+1$, but that is not consistent with $\ln e=1$.

  • $\begingroup$ Would be useful to clarify why this doesn't mean complex analysis is an inconsistent theory. $\endgroup$ – akkkk Aug 4 '12 at 16:10
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    $\begingroup$ @Auke: No. That would lead to a paraphrase of half a chapter's worth (or more!) of some complex analysis book, talking about branch points, Riemann surfaces, and I don't know what. This is not the purpose of this site. Anyway, the point of all this is that the union of complex analysis and the blind application of certain rules learned from real analysis is an inconsistent theory. $\endgroup$ – Harald Hanche-Olsen Aug 4 '12 at 17:50

From wikipedia on Euler identity

The identity is a special case of Euler's formula from complex analysis, which states that $e^{i x}=\cos x+i\sin x$. for any real number $x$.

Note $x$ should be real number.

$$e e^{i x} \neq e^{i x + 1} = e^{i(x-i)} = \text{undefined} $$

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    $\begingroup$ Euler's formula also holds for complex $x$. In fact, it is a great trick to define cos and sin for arbitrary arguments. $\endgroup$ – akkkk Aug 4 '12 at 16:07
  • $\begingroup$ Oh! so $e*e^{ix} = e^{ix + 1}$ is true? $\endgroup$ – Ankush Aug 4 '12 at 17:15
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    $\begingroup$ @Ankush: Yes. But please, please don't use the asterisk $*$ for multiplication! In mathematics, it usually means convolution. It's an abomination coming from programming languages, where it arose due to the lack of a more appropriate symbol in ASCII (or EBCDIC, or …). Use \cdot instead: $e\cdot e^{ix}$. $\endgroup$ – Harald Hanche-Olsen Aug 4 '12 at 17:41
  • $\begingroup$ ok :) I'm new to writing equations in $ $\endgroup$ – Ankush Aug 4 '12 at 18:27
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    $\begingroup$ @Ankush: it is called LaTeX. $\endgroup$ – akkkk Aug 4 '12 at 19:22

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