Number of functions from naturals to reals (${\aleph_1}^{\aleph_0}$)

Question

The number of functions from $\mathbb{N}$ to itself is $|\mathbb{N}|^{|\mathbb{N}|}={\aleph_0}^{\aleph_0}=\aleph_1$. The number of functions from $\mathbb{R}$ to itself is $|\mathbb{R}|^{|\mathbb{R}|}={\aleph_1}^{\aleph_1}=\aleph_2$. Finally, the number of functions from $\mathbb{R}$ to $\mathbb{N}$ is $|\mathbb{N}|^{|\mathbb{R}|}={\aleph_0}^{\aleph_1}=\aleph_2$. My question is whether the number of functions from $\mathbb{N}$ to $\mathbb{R}$ ($|\mathbb{R}|^{|\mathbb{N}|}={\aleph_1}^{\aleph_0}$) is $\aleph_1$ or $\aleph_2$. Thanks!

EDIT:

Please note that I assume the generalised continuum hypothesis.

You seem to have assumed the generalised continuum hypothesis — Henry, Jan 20 '21 at 13:25
According to W|A, there is no known simplified form. Also, the former simplifications work only if you're assuming CH/GCH — Prasun Biswas, Jan 20 '21 at 13:26

score 1 · Accepted Answer · answered Jan 20 '21 at 13:41

I will write $\mathfrak c = 2^{\aleph_0}$.
Compute $$\aleph_1^{\aleph_0} \ge 2^{\aleph_0} = \mathfrak c \\ \aleph_1^{\aleph_0} \le \mathfrak c^{\aleph_0} = \big(2^{\aleph_0}\big)^{\aleph_0} = 2^{\aleph_0\times\aleph_0}= 2^{\aleph_0} = \mathfrak c $$ therefore $$ \aleph_1^{\aleph_0} = \mathfrak c $$

Number of functions from naturals to reals (${\aleph_1}^{\aleph_0}$)

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