Question: * The matrix is $n\times n$
* The nth row is $[n^1,n^2,n^3....n^n]$

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I tried the first few examples:
$\begin{pmatrix}1\end{pmatrix}$ ,$\begin{pmatrix}1 & 1\\ 2 & 4\end{pmatrix}$,$\begin{pmatrix}1 & 1 & 1\\ 2 & 4 & 8\\3 & 9 & 27\end{pmatrix}$. They are obviously invertible.

But afterwards I got confused. Do I try to prove that the determinant keeps increasing, or should I try proving by induction?

Answer: 
This matrix is related to the Vandermonde's matrix. It is always invertible because the determinant is always non-zero.

There's an easy procedure to compute the determinant of the Vandermonde matrix.

When calculating the determinant of your matrix, pull out a common multiplier $k$ from each row (where $k$ is the row number). Then you are left exactly with Vandermonde's determinant. The rest is standard theory.

<https://en.wikipedia.org/wiki/Vandermonde_matrix>
Given $1,...,n$, all of the polynomials of degree $n$ that has this numbers as roots are:

$p(x)=k(x-1)(x-2)...(x-n)=k(\tilde{a}\_0+\tilde{a}\_1x+...+\tilde{a}\_{n-1}x^{n-1}+\tilde{a}\_n x^n), k\neq 0$

So we are saying that the linear system
$$\begin{cases} a\_0+ja\_1+j^2a\_2+...+j^na\_n=0\end{cases} \forall j\in \{1,...,n\}$$
Has this set of solutions $\mathcal{S}=\{(k\tilde{a}\_0,..,k\tilde{a}\_n)|k\in \mathbb{R}\}$. Notice that if we fix the value of the first component of the solution (for example $a\_0=1$), then the system has a unique solution, because $k\tilde{a}\_0=1 \Rightarrow k=\tilde{a}\_0^{-1}$(notice that $|\tilde{a}\_0|=n!\neq0$ ), so the unique solution would be $(1,\tilde{a}\_0^{-1}\tilde{a}\_2,..,\tilde{a}\_0^{-1}\tilde{a}\_n)$. We are saying that this system:
$$\begin{cases} 1+ja\_1+j^2a\_2+...+j^na\_n=0\end{cases} \forall j\in \{1,...,n\}$$
Has unique solution. We can easily rewrite the system as:
$$\begin{cases} ja\_1+j^2a\_2+...+j^na\_n=-1\end{cases} \forall j\in \{1,...,n\}$$
So we are saying that this square system has unique solution, that is to say that its coefficients matrix is non singular. And the coefficients matrix of this system is exactly the matrix of your problem.