In two dimensional space, the length of a vector is $$\sqrt{x^2+y^2}$$
In three dimensional space, the length of a vector is $$\sqrt{x^2+y^2+z^2}$$
How can one prove that in n th dimensional space the length of a vector is
$$\sqrt{d_1^2+d_1^2+\cdots +d_n^2}$$
where $d_n$ represents the n dimensional axis, like $n_1=x, n_2=y, n_3=z, etc$
Using the Pythagorean theorem for higher dimensions, one can show that the length of a point $\mathrm{P}(d_1,\ldots,d_n)$ from the origin in $\mathbf{R}^n$ is precisely equal to, $$\big\lVert\vec{\mathrm{OP}}\big\lVert=\operatorname{dist}(\mathrm{O},\mathrm{P})=\sqrt{\displaystyle\sum_{k=1}^n d_k^2}.$$