Show by induction that$$(n!)^2<2^{n^2} \quad \forall n \in \mathbb{N} $$
May I get some help? Please.
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As noted by user10354138, we should prove $k + 1 \leq 2^k$ for all $k$. That is, we should prove that $k < 2^k$ for all $k$. If we know that $k < 2^k$, then we also know that $k + 1 \leq 2^k$, and thus $(k + 1)^2 \leq (2^k)^2 = 2^{2k} < 2^{2k + 1}$.
There are multiple ways of doing this. One is using a combinatorial proof. Note that $k = |\{1, 2, ..., k\}|$, and $2^k = |P(\{1, 2, .., k\})|$. Using Cantor's diagonalisation argument that $|S| < |P(S)|$ for all sets $S$ gives us that $k < 2^k$.
Another way of doing this is by another induction. The base case is $k = 0$, in which we see that $0 < 2^0 = 1$ holds.
The inductive step allows us to assume that $j < 2^j$ and tasks us with showing that $j + 1 < 2^{j + 1}$. We see that $j + 1 < 2^j + 1 \leq 2^j + 2^j = 2^{j + 1}$, thus completing the inductive step.
A third way of doing this is by using a certain general fact about increasing functions.
That lemma is as follows:
Suppose $f : \mathbb{N} \to \mathbb{N}$ is a function such that for all $n \in \mathbb{N}$, $f(n + 1) > f(n)$. Then for all $n \in \mathbb{N}$, we have $f(n) \geq n$.
This lemma can be proved quite simply by using induction. One then applies the lemma to $f(x) = 2^x - 1$ to show that $n \leq 2^n - 1$ for all $n$; hence $n < 2^n$ for all $n$.
So you want to prove $(k+1)^2<2^{2k+1}$.
It suffices to prove $k+1\leq 2^k$, since squaring both sides (as $k+1>0$) gives $(k+1)^2\leq 2^{2k}<2^{2k+1}$.
But binomial expansion $2^k=(1+1)^k=1^k1^0+k1^{k-1}1^1+\dots$ is a sum of $k+1$ nonnegative summands, and just summing the first two summand gives $2^k\geq 1+k$, as long as $k+1\geq 2$ which is indeed what we have in inductive hypothesis.