How to prove the following inequality when k > 10?

Let's say we have $n \in \Bbb N$ and $n \geq 10$, we need to prove the following with proof to back down:

$2^n \geq 100n$

I've tried the following:

• First I proved the first term of the inequality $n_ 0 = 10$, which holds

• Then I set $k$ to be another term, and $k \gt n_0$, so $k \gt 10$

• I attempted to prove that $k+1$ holds too.

I tried to prove $k+1$ by replacing $k$ with $k+1$ in the expression in order to get back to the expression with $k$, as such:

$2^{k+1} \gt 100(k+1)$

$2^{k+1} \gt 100k + 100$

But we have $2^k > 100k$, so:

$2^{k+1} - 2^k \gt 100$

$2^{k} > 100$

And then I don't know how to do it. Could anyone help?

By induction.
Case $n=10$:
We have that $2^{10}=1024 \geq 100 \cdot 10=1000$.
General case:
Supose that $k \geq 10$ and $2^{k} \geq 100 \cdot k$ then we have that:
$k+1 \geq 10$ and:
$2^{k+1}=2^{k} \cdot 2 \geq 100 \cdot k \cdot 2 = 100 \cdot (k+k) \geq 100 \cdot (k+1)$