The question is on induction. $$2^{n+1}<1+(n+1).2^{n}$$.
I proved it but need a verification at the last step of the proof.
For $n=k$ $$2^{k+1}<1+(k+1).2^{k}$$
For $n=k+1$ $$2^{k+1}.2^2<(1+(k+1).2^{k})2^2\\ 2^{k+2}<(2+(2k+2).2^{k})\\ < 1+1+(k+1)2^{k+1}\\ <1+(k+2).2^{k+1}$$.
Any suggestion for the last step please. I am not satisfied with the last two steps. Any help is greatly appreciated.
For the base case, we test with $n = 1$, and note that $$ 4 = 2^2 < 1 + (1 + 1)2^1 = 1 + 4 = 5. $$ Now, assume it holds until $n = k$. That is, assume that we have that $$ 2^{k + 1} < 1 + (k + 1)2^{k}. $$ Now, for the $n = k + 1$ case, try multiplying by 2 on both sides instead of $2^2$. We get \begin{align*} 2^{k + 1} \cdot 2 &< (1 + (k + 1)2^{k}) \cdot 2 \\ 2^{k + 2} &< 2 + (k + 1)2^{k + 1}. \\ \end{align*} Now we can use the inequality that \begin{align*} 2 + (k + 1)2^{k + 1} &= 1 + 1 + 2^{k + 1} k + 2^{k + 1} \\ &< 1 + 2^{k + 1} + 2^{k + 1} + 2^{k + 1}k \\ &= 1 + 2(2^{k + 1}) + k(2^{k + 1}) \\ &= 1 + (k + 2)2^{k + 1}. \end{align*} Combining all these inequalities gives that $$ 2^{k + 2} < 2 + (k + 2)2^{k + 1}, $$ as desired. Thus, by the induction hypothesis, the result holds for all $n$.