Prove by mathematical induction $n < 2n$ Prove the following by mathematical induction

$n &lt; 2n$, for all positive integer $n$.

This is what I have done:

Step 1:
$n=1$: $1 &lt; 2$

Step 2:
$k &lt; 2k$

$n=k+1$: $(k+1) &lt; 2(k+1)$

$k + 1 &lt; 2k + 1 &lt; 2k + 2 = 2(k+1)$

Hence $P(k+1)$ is true whenever $P(k)$ and since $P(1)$ is true.

I didn't write all necessary assumptions but can anyone help me to check if my method is correct or if it needs improvements. Thank you.
We have to prove that $$n+1&lt;2(n+1)$$ , adding $1$ on both sides of $$n&lt;2n$$ we get
$$n+1&lt;2n+1$$ and $$2n+1&lt;2(n+1)$$ so the proof is finished.
We have to prove that $$n+1&lt;2(n+1)$$ , adding $1$ on both sides of $$n&lt;2n$$ we get
$$n+1&lt;2n+1$$ and $$2n+1&lt;2(n+1)$$ so the proof is finished.