How do I prove that if $n$ is greater or equal to $4$, then $(2n)!$ is greater than or equal to $10^n$? I used the method of induction to prove this.
For the basis step, when $n=4$, the statement is $[2(4)]!≥10^4$ which is the same as $40320≥10000$ which is true.
Next, for the inductive step, we show $\mathrm{S}_k$ implies $\mathrm{S}_{k+1}$ for some integer $n=k≥1$.
By assuming that $(2k)!≥10^k$ is true. Therefore, we need to show $(2k+2)!≥10^{k+1}$ for $n=k+1$. First, $10^{k+1}=10\cdot10^k$. By the induction hypothesis, $(2k)!≥10^k$, so $10\cdot(2k)!≥10\cdot10^k$.
Since $k≥4$, we know that $2(k+1)≥10$.                So, $2(k+1)\cdot(2k)!≥10\cdot10^k$ by substitution. This simplifies to $(2(k+1))!≥10\cdot10^k$ which equals $(2(k+1))!≥10^{k+1}$ which we needed to show.
I know my expansion from $2(k+1)(2k)!$ to $(2(k+1))!$ is wrong, but I don't know what the correct way to get to $(2(k+1))!$ from $2(k+1)(2k)!$.

With induction but more clear:
You have proved the base case, so let's continue.
Let $(2n)!\ge 10$. Then:
$$(2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!=(4n^2+6n+2)(2n)!\ge (4n^2+6n+2)\cdot 10^n$$
Clearly $4n^2+6n+2 \ge 10$ for $n \ge 1$ so:
$$(2(n+1))! \ge (4n^2+6n+2)\cdot 10^n \ge 10\cdot 10^n=10^{n+1}$$
$$(2(n+1))! \ge 10^{n+1}$$
And we have proved the inductive step $\blacksquare$
Just to clarify, you were actually almost there. We know that $2(k+1)\cdot (2k)! = (2k+2) \cdot (2k)! \le (2k+2)(2k+1) \cdot (2k)! = (2(k+1))!$
If $2(k+1)\cdot (2k)! \ge 10^{k+1}$ and $(2(k+1))! \ge 2(k+1)\cdot (2k)!$ then $(2(k+1))! \ge 10^{k+1}$
