Find all $n$ for which $2^n \ge (n+1)^2$ 
Find all of the elements of $X= \{ n \in \mathbb N: 2^n \ge (n+1)^2\}$

Could someone give me a hint to nudge me in the right direction?
 A: Find the smallest element $s$ in $X$ by hand. Then use induction to prove that for all $k\geq s$ : $k\in X\implies (k+1)\in X$.
A: $\color{green}{2^0=1\ge(0+1)^2=1}$ ?*
$\color{red}{2^1=2\ge(1+1)^2=4}$ ?
$\color{red}{2^2=4\ge(2+1)^2=9}$ ?
$\color{red}{2^3=8\ge(3+1)^2=16}$ ?
$\color{red}{2^4=16\ge(4+1)^2=25}$ ?
$\color{red}{2^5=32\ge(5+1)^2=36}$ ?
$\color{green}{2^6=64\ge(6+1)^2=49}$ ?
$\color{green}{2^7=128\ge(7+1)^2=64}$ ?
$\color{green}{2^8=256\ge(8+1)^2=91}$ ?
$\color{green}{2^9=512\ge(9+1)^2=100}$ ?
...
*whether $0$ is considered natural or not is a matter of convention.
A: It is interesting to find the values at which strict equality $2^n=(n+1)^2$ occurs, which may be real numbers.
This is a transcendental equation which cannot be solved analytically.
Let us first take the square roots, to get a linear RHS:
$$\sqrt2^n=n+1.$$
And let us derive to find extrema:
$$\ln\sqrt2\ \sqrt2^n=1.$$
The single solution, $-\frac{\ln\ln\sqrt2}{\ln\sqrt2}$, lies between $3$ and $4$, so that $2^n-(n+1)^2$ decreases from $0$ (value $0$) to $3$ (value $-8$) then increases after $4$ (value $-9$).
So there is a root at $n=0$, and another at some $n>4$.
At this stage, there is little better to do than trial and error with increasing $n$ values. Being pessimistic, we will use exponential search first, i.e. doublings of $n$.
$$f(8)=175\ge0.$$
Now the solution is bracketed by $[4,8]$ and we will continue with dichotomic search:
$$f(6)=15\ge0,$$
$$f(5)=-4<0.$$
And we are done, $X=[0]\cup[6,+\infty[$.

Assume now that we need to solve for $2^{n-1000}\ge(n+1)^2$ instead. Following the same procedure, we find that the function increases for $n\ge 1003$, $f(1003)=-1008008$.
Then
$$f(2006)\gg0$$
$$f(1504)\gg0$$
$$f(1253)\gg0$$
$$f(1128)\gg0$$
$$f(1065)\gg0$$
$$f(1034)=17178797959\ge0$$
$$f(1018)=-776217<0$$
$$f(1026)=66054135\ge0$$
$$f(1022)=3147775\ge0$$
$$f(1020)=6135\ge0$$
$$f(1019)=-516112<0.$$
$X=[1020,+\infty[$.
(At some stage, switching to the secant method can be an advantage.)
A: Hint: You can prove by induction on $n$ that $2^n \geq n^2$ whenever $n\geq 4$. Then note that $(n+1)^2 = n^2 + 2n + 1$.
Edit: Just to keep the record straight, in reference to comments above, it is true that $Y = \{n \in \mathbb N\mid 2^n \geq m^2\} = \{n\in \mathbb N\mid n\geq 4\}$.  However, to satisfy membership in the posted inequality, $$X = \{n \in \mathbb N\mid 2^n \geq (n+1)^2\}=\{n\in \mathbb N\mid 2^n \geq n^2 + 2n + 1\}$$ requires that $n \geq 6$. So we need to reject $4, 5$, since $2^4 = 16 \lt (4+1)^2 = 25$, and $2^5 = 32 \lt (5+1)^2 = 36$. So $2^n \geq (n+1)^2$ is a stricter requirement on $n \in \mathbb N$ than is $2^n \geq n^2$.
