Prove for all $m, n \in \mathbb N$: $[1 + 3 +\cdots + (2n -1)]^m = n^{2m}$ I have this:
Prove for all $m, n \in \mathbb N$: 
$$[1 + 3 + \cdots + (2n - 1)]^m = n^{2m}$$
For $n = 1: 1 = 1^2$, hence P(1) is true.
Let $N \in \mathbb N$ be given and assume: 
$$[1 + 3 + \cdots + (2N - 1)]^m = N^{2m}$$
For $n = N + 1$: 
$$[1+3+\cdots+(2N-1)+(2(N+1)-1)]^m = (N+1)^{2m}$$
But I don't know what to do know or did I already made a mistake?
 A: Apart from the standard proof by induction for the sum of the first odd numbers $1+3+\cdots +(2n-1)$, I'm pretty sure that you are all aware of the beautiful "picture proof" :

A: Prove it for $m=1$ using induction on $n$. Then exponentiate.
A: Note that
$$1+2+\cdots+n=\frac{n(n+1)}{2}$$
Thus,
$$1+3+\cdots+(2n-1) = (1+2+\cdots+(2n))-(2+4+\cdots+(2n))$$
$$=(1+2+\cdots+(2n))-2(1+2+\cdots+n)$$
$$=\frac{(2n)(2n+1)}{2}-2\cdot\frac{n(n+1)}{2}$$
$$=n^2$$
Therefore,
$$\{1+3+\cdots+(2n-1)\}^m=(n^2)^m=n^{2m}$$
A: Umberto has the key, and I feel like typing. I hope I am not spoiling someone's homework.
$\textbf{Theorem.} \, \forall \, m,n \in \mathbb{N}, (1+3+...(2n-1))^m=n^{2m}.$
$\textbf{Proof.}$ We begin by proving that $(1+3+...+(2n-1))=n^{2},$ 
by induction on $n$, and then conclude that $(1+3+...+(2n-1))^m=n^{2m}$ by properties of exponents.
For for the base case let $n=1$. Then
$$(2 \cdot 1-1)=1=1^2.$$
Assume the series holds for $n=k$. Then
$$(1+3+...+(2k-1))=k^2.$$
For the inductive step, we let $n=k+1$ to show that
$$(1+3+...+(2k-1)+(2k+1))=(k+1)^2.$$
On the left side of the equality we have,
$$
\begin{align*}
(1+3+...+(2k-1)+(2k+1)) &= (1+3+...+(2k-1)) + (2k+1) \\
                        &= k^2 + (2k+1) \\
                        &= k^2+2k+1 \\
                        &= (k+1)^2.
\end{align*}
$$
Thus we have shown that
$$(1+3+...(2n-1))=n^{2}.$$
Now by exponentiation on both sides by $m$,
$$\qquad\qquad (1+3+...(2n-1))=n^{2} \Rightarrow (1+3+...(2n-1))^m=n^{2m}. \qquad\qquad \blacksquare$$
Please feel free to refine this, I no longer feel like typing.
