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The "rule" $a^{bc}=(a^b)^c$ simply does not hold for complex numbers. Indeed, it doesn't even hold for all real numbers (consider $a=-1$ and $b=2$ and $c=\frac12$). This is a great example of how, as mathematicians, we need to remember more than just the equations we learn—we need to remember the precise conditions under which the equations hold.

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According to page 8, Problem 25 in , the equivalent problem was published by Vasile Cirtoaje on Gazeta Matematica, Seria B, No. 10, 1991. Five solutions are given in . Here is one of them: Let $r = \frac{n-1}{n}$. By AM-GM, we have, for $i=1, 2, \cdots, n$, $$\sum_{j\ne i} x_j^r \ge (n-1) \left(\prod_{j\ne i} x_j\right)^{r/(n-1)} = (n-1) x_i^{r - 1}$$ ...

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Hint: by definition, $b^c = \exp(c \log b)$. Use all values of $\log b$.

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If, when $a>0$, you define $a^z$ as $\exp\bigl(z\log(a)\bigr)$, where $\log(a)$ is the only real logarithm of $a$, then indeed we always have $1^z=1$. Otherwise, it's up to you to tell us how you are defining $1^z$. However, note that the set of all logarithms of $1$ is not $\{0,2\pi i,4\pi,i,6\pi i,\ldots\}$; it's $\{0,\pm2\pi i,\pm4\pi i,\pm6\pi i,\... 3 To read aloud I would say, "x to the power, pause, n minus 1", and "x to the power n, pause, minus 1". The position of the pause is used to communicate the implied brackets. If this is not clear enough, then "x to the power open bracket n minus 1 close bracket". 3 One can use Lagrange multipliers for: $$L = \sum_{k=1}^{n} 1/(n-1 + x_{k}) - \lambda(\prod_{k=1}^{n}x_{k} - 1)$$ One has: $$\frac{dL}{dx_{k}} = -\frac{x_{k}}{(n-1+ x_{k})^{2}} -\lambda\prod_{j\neq k}x_{j}$$ And when this is zero, one can multiply both sides by$x_{k}$to get: $$\frac{x_{k}^{2}}{(n-1+ x_{k})^{2}} =\lambda\prod_{k}x_{k}$$ Because of condition ... 3 The question finds an immediate answer by noting that$b^h=e^{\ln b^h}=e^{h\ln b}$. Therefore $$L(b) = \lim_{h\to 0}\frac{b^h-1}{h}=\lim_{h\to 0}\frac{e^{h\ln b}-1}{h} = \ln b\lim_{t\to 0}\frac{e^t-1}{t}=\ln b.$$ I have multiplied and divided by "$\ln b$" the fraction and substituted$h\ln b = t$to get the result. 3 I cannot confirm your counterexample. Here are the details so nobody else has to duplicate this effort. Following the method from Brian Moehring's answer, we want to find a value of$n$such that$\log_5(3^n - 1), \log_7(3^n - 1), \log_{13}(3^n - 1)$all have fractional parts close to$1$. If their integer parts are$n_5, n_7, n_{13}$, it then suffices to ... 3 In binary, this is $$1+1+10+100+...+1\underbrace{00...00}_{n-2\text{ times}}=1+\underbrace{11...11}_{n-1\text{ times}}=1\underbrace{00...00}_{n-1\text{ times}}$$ which in decimal is$2^{n-1}$. 3 What you suggest basically seems to work - I'm going to change the leading value into$n$rather than$n-1$for ease of reading: First,$2^1 = 1+1$and$2^2 = 1+1+2 = 1+2^0+2^1$Then using induction, assuming that the summation holds for lower$n$, we can use your observation$2^n = 2^{n-1} + 2^{n-1}$and change the first$2^{n-1}$into the summation form, ... 2 To make the problem more symmetric, let$x=t+\frac 12$and expand the function as Taylor series around$t=0$. You will have $$f(t)=1+\sum_{n=1}^p a_n t^{2n}$$ where the$a_n$'s are polynomials of degree$2n$in$k=\log(2)$$$a_1=\left\{2,-\frac{13}{4},\frac{1}{2}\right\}$$ $$a_2=\left\{\frac{15}{4},-\frac{1607}{192},\frac{439}{96},-\frac{23}{24},\frac{1}{24}\... 2 Some thoughts Let me show how to use bounds for the case 0 < x < \frac{1}{10}. Denote F = W(2\mathrm{e}x)^{2x} and G = W(2\mathrm{e}(1-x))^{2(1-x)}. We need to prove that x^F + (1-x)^G \le 1. Fact 1: If u > 0 and 0 \le v \le 1, then u^v \ge \frac{u}{u + v - uv}. (Note: By Bernoulli inequality, (\frac{1}{u})^v=(1+\frac{1}{u}-1)^v\leq 1 ... 2 One approach is to define e as the unique number b for which$$ \lim_{h \to 0}\frac{b^h-1}{h}=1 \, . $$It then directly follows that e^x is its own derivative. Then, since b^h=e^{h\log(b)}, it can be shown that the above limit is equal to \log(b) in general. (See Gauge_name's answer.) 2 Hint: x^{13}=x^{12} x = (x^2)^6 x 2 x=9/4 is a solution \sqrt(9/4)=3/2 3/2-1=1/2 9/4^{\sqrt(9/4)-1}=\sqrt(9/4)=3/2 2 EDIT: As pointed out, this answer contains a flawed application of Cauchy-Schwarz. In particular, the L.H.S. of the inequality can be zero, while the R.H.S. is positive, an absurdity. I tried to remedy this flaw but ultimately failed. Hence, I have come up with a new approach that proceeds by a "smoothing principle" argument. Any feedback on the ... 2 You're right that every non-zero complex number has infinitely many logarithms, but it is possible that exponentiation will make all these values equal. For instance, let z= r e^{i \theta}, then the possible values of \log(z) are \log(r) + i \theta + 2 \pi i k for k \in \mathbb{Z}. Now you would like \exp(w\log z) to be a well-defined function ... 2 The case x_1=x_2=...=x_n=1 is trivial, and for n >2, is the only case when equality holds. We prove this below. Suppose the x_is are not all 1. Then, there exists i,j \in \{1,2,..,n\}, i \neq j, such that x_i<1<x_j. Replace the pair (x_i \ ,x_j) with (x_i'\ ,x_j') , such that:$$x_i'=1, \ x_j'=x_ix_j.$$x_i' and x_j' have the ... 2 This statement holds only for s>0. For s\ne0 the anti-derivative of x^{-s-1} is -\frac{x^{-s}}s. Applying the limits,$$\frac s{-s}[x^{-s}]_{p^n}^\infty=p^{-ns}$$since \infty^{-s}=1/\infty^s=0 for positive s. 2 Here is one way to solve it: (2n+1)\pmod 7\equiv 1,3,5,0,2,4,6,\cdots\ and cycling. 2^{4n+5}\pmod 7\equiv 4,1,2,\cdots\ and cycling. Both can be proved by induction on n. 1\times 3\equiv 2\times 5\equiv 4\times 6\equiv 3\pmod 7\ verify there are no others. Finally can you find n for which the conditions are reunited ? (hint, Chinese theorem).... 2 Update: I found that this solution provides a neat way in general to solve this kind of problems. I will apply the method here:$$(2n+1)2^{4n+5} \equiv 3 \pmod 7 \\ \iff g(n)=(2n+1)2^n\equiv (2n+1) 2^{4n+6} \equiv 3\cdot 2 \equiv -1 \pmod 7$$Suppose n=3k+i, i=0,1,2, then$$-1 \equiv g(3k+i) = (6k+2i+1)2^{3k+i} \equiv (-k+2i+1)2^i=g(i)-k\cdot 2^i \pmod 7\\ ... 2 Let me suggest 4 ways:$ a^{n}-b^{n}=(a-b)(a^{n-1} + ba^{n-2}+ ... + b^{n-1}) \Rightarrow 2^{n} = 2^{n} - 1 + 1 = 1+2+ ... +2^{n-1} +1 $Considering all subsets from set with$ n>1 $elements and considering map between subsets and binary numbers Mathematical Induction Newton's binomial theorem 2 It's because it's a geometric progression. Since $$1+p_i+{p_i}^2+··· +{p_i}^{a_i} = \dfrac{{p_i}^{a_i+1}-1}{p_i-1}.$$ you just add an extra one at the start of the progression. 1 You forgot that$(-1)^{2n} = 1$for positive$n$. (The exponent is even). You have different cases for$x^y = 1$: $$x \not= 0, y = 0$$ $$x = 1, y = \text{any real number}$$ $$x = -1, y = \text{even number}$$ You missed the last case. 1 I will give only the two other solutions the you missed. It is trick. Look carefully for the equation. RHS equal 1. if$x^2-13x+42$is even number. then$x^2-7x+11=-1$is a solution. By factoring the last equation , we have$x=3$and$x=4$. So, the set of solution is$\{2,3,4,5,6.7\}$1 Welcome to MSE!$\sqrt 4$is an integer, not irrational. In your argument,$4b^2=a^2$, so$2^2b^2=a^2$and there is no way to find a contradiction. So if the number is a perfect square, like 4, 9, 16, the proof of contraction fails as it should. 1 Partial answer : Remarking that the function$f(x)=\frac{e^x}{e^x(n-1)+1}$is concave on$(-\ln(n-1),\infty)$and rewriting the inequality like : $$S=\sum_{i=1}^{n}\frac{e^{x_i}}{e^{x_i}(n-1)+1}\leq 1$$ We apply Jensen's inequality to get the value$1$since$e^{\sum_{i=1}^{n}x_i}=1$and$x_i\in(-\ln(n-1),\infty)$1 For$x^{n-1}$:$x$raised to the difference of$n$and$1$. And for$x^n-1$: The difference of$x$, raised to$n$, and$1$. 1 For$x^{n-1}$,$x$whole raised to$n-1"$looks great. For$x^n - 1$, I'd suggest$1$less than$x^n"$1 Let's substitute,$x=y^2$, $$x^{\sqrt{x}-1}=(y^2)^{y-1} = y^{2y-2} = \frac{3}{2}$$ $$\implies y^{y-1} = \sqrt{\frac{3}{2}}$$ This is still an implicit equation but we got rid of the$\sqrt{x}\$. Here is the Numerical Solution by Wolfram. This problem cannot be solved analytically using elementary operations. You need to follow some algorithm like [...

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