We're working in $\sf{ZF}$ set theory, we assume $n$ is a finite cardinal, and the set $X$ is not finite. My question is: what's the smallest cardinal $k$ for which $\sf{ZF}$ proves $|X^n| \leq |k^X|$? We know that such $k$ must exist and be finite, since $|X^n|\leq |2^X|^n=|(2^n)^X|$ proves $k\leq 2^n$. The case $n=0$ is trivial, $k_0=1$. The case $n=1$ isn't much harder, $k_1=2$. The analogous question for $\sf{ZFC}$ is also trivial, $k_n=2$ for all $n>0$, but without Choice the problem seems rather difficult even for $n=2$.
I was able to prove the upper bound $k_n\leq n+1$ for all $n$, and I suspect this is optimal. In particular, I suspect that for each $n$, it's consistent that there exists infinite $X$ for which $|X^n|\not\leq |n^X|$. More strongly, I suspect it's consistent that there exists $X$ for which all $n$ have $|X^n|\not\leq |n^X|$. Unfortunately I'm not very familiar with forcing techniques, so I don't know how to prove this. It might be possible to establish a nice lower bound by considering the case where $X$ is an amorphous set, but I couldn't figure out how to do it. I would accept any answer that at least finds $k_3$.
To prove that $k_n\leq n+1$, I'll start by proving $|X^n|\leq n^n|(n+1)^X|$. Consider a function $f:n\to X$. We use $f$ to define the functions $p:X\to (n+1)$ and $g:n\to n$ like so. $$p(x):=\begin{cases}n &: x\notin f[n] \\ \min\{j<n : x=f(j)\} &: x\in f[n]\end{cases}$$ $$g(j):=p(f(j))$$
Notice that the map $f\mapsto (g,p)$ sends $X^n\to n^n\times (n+1)^X$. This map is moreover an injection, since we have $p^{-1}[g(j)]=\{f(j)\}$ for all $j<n$. The intuition behind this strategy is that the function $p$ injects the image of $f$ into $n$, and then we use this numbering to compress $f$ into a function $g:n\to n$. Since $p$ can be used to decompress $g$ back into $f$, we have an injection. From these arguments, it follows that $|X^n|\leq n^n|(n+1)^X| \leq |(n+2)^X|$ and therefore $k_n\leq n+2$.
For $n>0$, the above strategy can be further improved to give $k_n\leq n+1$. To do this, let $M=(n+1)\times(n+2)$, and find an injection $\iota:M\to X$, which is possible since $X$ is infinite. For each $j<n+1$, define the sets $B_j=\{\iota(j,i) : i<n\}$ and $I_j=\{\iota(j,i) : n\leq i<n+2\}$, so that the collection of all $I_j$ and $B_j$ are pairwise disjoint, and each $|B_j|=n$ and $|I_j|=2$. Using the same map $f\mapsto (g,p)$ as before, we notice that there necessarily exists $j<n+1$ for which we have the image $p[B_j\cup I_j]=\{n\}$. Relying on that fact, we define the map $(g,p)\mapsto q\in (n+1)^X$ as follows. $$J:=\min\{j<n+1 : p[B_j\cup I_j]=\{n\}\}$$ $$q(x):=\begin{cases} p(x) &: x\notin (I_J\cup B_J) \\ 0 &: x\in I_J \\ g(i) &: x\in B_J\land x=\iota(J,i) \end{cases}$$
To recover $(g,p)$ from $q$, first we recover $J<n+1$ as the unique number satisfying $q[I_J]=\{0\}$. Now we easily recover $g$ via $g(i)=q(\iota(J,i))$, and for $p$ we have $p(x)=q(x)$ when $x\notin (I_J\cup B_J)$ and $p(x)=n$ otherwise. The recoverability of $(g,p)$ from $q$ proves this is an injection. The intuition behind this strategy is that $p$ leaves a lot of empty space on which $p(x)=n$ is constant, and we can fit enough information in that space to encode $g$. We partition a subset of $X$ into some blocks $B_j$ and indicators $I_j$, and based on the structure of $p$, we're guaranteed that some block/indicator pair will be vacant. We choose one of these vacant block/indicator pairs, encode $g$ inside the block $B_j$, and enable the indicator $I_j$ so we know where to look to decode $g$. By composing our injections, we can send $f\mapsto q$ to prove $|X^n|\leq |(n+1)^X|$ and thus $k_n\leq n+1$ for all $n>1$. A similar strategy should prove $m|X^n|\leq |(n+1)^X|$ for all finite $m$, which is an interesting albeit unrelated result.