Difference between order of convergence and order of consistency? I understand that, as the time step is reduced:

Consistency: The local error must approach zero.
Convergence: The global error must approach zero.

But why are both the order of convergence and consistency of explicit Euler = 1, if the global error is $O(h)$ while the local error is $O(h^{2})$?

More details: The Euler method is used to approximate an initial value ODE by time stepping using the derivative.

For example, for ODE form:
$y'(t) = f(t,y(t)), \qquad \qquad y(t_0)=y_0 $

The solution at point $n$ is:
$y_{n+1} = y_n + hf(t_n,y_n)$

This formula can be obtained from the Taylor expansion, where $h$ is your time step:

$y(t + h) = y(t) + hy'(t) + \frac{h^{2} y''(t)}{2!} + ... $

So the local error is of order $O(h^{2})$. Now the global error is one order less which is why Euler is called a first-order method. So my question is: Why is the order of consistency also 1, even though the local error is of order 2?
A quick and dirty answer: by definition.

Let's use the definition from Wikipedia.
Let $\delta^h_{n+k}$ be the local error.
A method has order $\textbf{p}$ if
$$\delta^h_{n+k} = O(h^{\textbf{p+1}}) \text{ as } h \to 0.$$

Notice $p$ and $p+1$ in the definition. Since the explicit Euler method has local error of order $O(h^2)$, $p+1 = 2$, hence $p = 1$.

On the other side, if you want to know the reason behind such definition, then I think it is about showing you the part of the solution's Taylor polynomial which the method reproduces correctly (by giving you the degree of the last correct term).