Let $f(n) = n^{(n+1)}$ can we say that $f(n)$ is $O(n^n)$?
To solve this problem I am using the Big-Oh definition:
We must find $c_1$ and $N_0$ such that $f(n) \leq C_1 \times n^2$ for $n \geq N_0$
My Attempt:
I say that we cannot say that $f(n)$ is $O(n^n)$ because $n^{(n+1)} $ will always be greater than $n^n$ for $(n \geq 1)$.
$ n^{(n+1)} \leq n^n$ for $(n \geq 1) \rightarrow$ This is false
Am I on the right track? I'm not sure if what i've written is enough of a proof.
EDIT: I've gotten a little further
I let $f(n) = n^{n+1}$ and $g(n)=n^n$.
Now, $N_0 \geq 1$ and $C_1 = 2$
Now, $n^{n+1} \leq 2 \times n^n$ for $n \geq 1 \rightarrow$ This is false
However,
I let $f(n) = n^{n+1}$ and $g(n)=n^n$.
Now, $N_0 \geq 1$ and $C_1 = 2$
Now, $n^{n+1} \leq 2 \times n^n$ for $n \geq 2 \rightarrow$ This is True
Is my truth enough to be represented as a proof?
Edit # 2
$n^{n+1} \leq C\times n^n$
$n^n \times n^1 \leq C \times n^n$
$n^1 \leq c$
Therefore, $N_0 = 1$ and $C_1 \geq 1$
$n^{n+1} \leq 1 \times n^n$ for $n \geq 1$
I also have another problem: Let $f(n) = n!$. Prove $f(n)$ is $O(n^n)$
My Attempt:
I feel as if the answer is simply:
$n! \leq n^n$ for $(n \geq 1)$
I can prove this using a graph, but do I need to show more math?
The blue line is $n^n$, and the red line is $n!$
