Q. Prove that $2^n>n for all positive integers n.$

Please explain the last line of solution

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$$\begin{gathered} We have P\left(n\right): 2^n > n, which is true for n = 1, as shown in the example in the picture. We have to prove that P\left(n\right) is true for any positive integer \text{'}n\text{'}, i.e, n\ge 1. \\ Let us assume that P\left(n\right) is true for some n = k (k \ge 1). \\ Then, 2^k > k ...\left(i\right) \\ Now, we have to prove that P\left(n\right) is true for n = k + 1. \\ Multiplying both sides of inequality \left(i\right) by 2, \\ 2.2^k > 2k ...\left(ii\right) \\ Here, 2k = k + k \\ \because k \ge 1 \\ \therefore k + k \ge k + 1 \\ \Rightarrow 2k \ge k + 1 ...\left(iii\right) \\ From \left(ii\right) and \left(iii\right), we have \\ 2.2^k > 2k \ge k + 1 \\ \Rightarrow 2^{k + 1} > k + 1 \\ Thus, by principle of mathematical induction, we have proved thet P\left(n\right) is true for any positive integer \text{'}n\text{'}. \end{gathered}$$

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