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could someone check my answer to see if its correct or if i missed something. Thank you.

Problem: $2^n < (n+1)!$ for all integers $n \ge 2$ (Use Induction)

Base Case: Show that $n = 2$ is true $2^2 < (2+1)!$ = $4<6$

Hypothesis: Assume that $ 2^k<(k+1)!$ is true for some int $k$ in $n$ and $k \ge 2$

Inductive step: Show that $n=k+1$ is also true \begin{align}2^{k+1} &=2^k \cdot 2^1\\&<(k+1)!2\\&<(k+2)(k+1)!\\ 2^{k+1}&<[(k+1)+1]! \end{align} Since the statement is true for $n=2$ and true for $n=k$, means that $n=k+1$ is also true and the statement is also true for all $n \ge 2$.

Siong Thye Goh
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user3365751
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1 Answers1

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It seems fine.

My minor comment is that you might want to make things clearer.

  • $2^2 < (2+1)! \color{red}= 4<6$

You want to write it at a different line. $$2^2 < (2+1)!$$ $$\iff 4< 6$$

  • write "$k$ in $n$" as $k \in \mathbb{N}$.
Siong Thye Goh
  • 153,832