Note that since $x_i$'s are positive reals with sum $1$, they are all strictly less than $1$, so $1-x_i$ is positive. Now note that by weighted AM-GM inequality $$x_i^{1-x_{i+1}}=1^{x_{i+1}}\cdot x_i^{1-x_{i+1}}\le 1\cdot x_{i+1}+x_i\cdot(1-x_{i+1})=x_i+x_{i+1}-x_ix_{i+1}<x_i+x_{i+1}.$$Summing this over all $i\in\{1,\cdots ,n\}$ (taking $x_{n+1}=x_1$) we see that $$x_1^{1-x_2}+x_2^{1-x_3}\cdots+x_{n-1}^{1-x_n}+x_n^{1-x_1}<2(x_1+\cdots +x_n)=2.$$This completes the proof. $\square$ Comment from Anish Kulkarni(test-solver) The key inequality can be rephrased with Bernoulli instead of Weighted AM-GM as we get $x_i^{1-x_{i+1}}= (1-(1-x_i))^{1-x_{i+1}}\le 1-(1-x_i)(1-x_{i+1})\le x_i+x_{i+1}-x_ix_{i+1}$

We do the only thing to deal with variables in exponents, WAMGM! By Weighted AM-GM, we have $x_{i-1} + x_i - x_ix_{i-1} = x_{i}(1) + (1-x_{i})(x_{i-1}) \ge x_{i-1}^{1-x_i}$. Summing, we have $\sum x_{i-1}^{1-x_i} \le 2 \sum x_i - \sum x_ix_{i-1} < 2$, as desired. $\blacksquare$

Here is my solution: https://calimath.org/pdf/IndiaEGMOTST2022-1.pdf And I uploaded the solution with motivation to: https://youtu.be/uP9YosvfC_E

Thanks,Cali.Math!

By Bernoulli Inequality, \[LHS=\sum{((x_i-1)+1)^{1-x_{i+1}}}\leq \sum{(1-x_{i+1})(x_i-1)+1}=2\sum{x_i}-\sum{x_ix_{i+1}}<2\]As desired.$\blacksquare$

Note that from WAMGM, $$x_i^{1-x_{i+1}}1^{x_{i+1}} \le x_i + x_{i+1} -x_ix_{i+1} < x_ix_{i+1}.$$Summing over $i$ yields the conclusion.