Don't mess up with the e`s!

Step 1:

Let $f(x) = e^{e^{e^x}}$

$f^{\prime} (x) = e^{e^{e^x}} e^{e^x} e^x\\$

Step 2: Prove that the $n^{th}$ derivative of $f$ has the following form, for $n \geq 1$

$f^{(n)} (x) = \displaystyle \sum_{\substack{1 \leq i \leq n \\ 1 \leq j \leq n }} c_{nij} e^{e^{e^x}} (e^{e^x})^i (e^x)^j$

Prove by mathematical induction:

For $n=1$ ,

$c_{111} = 1$

Assume the statement is true for $n=k$ ,

$f^{(k)} (x) = \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} e^{e^{e^x}} (e^{e^x})^i (e^x)^j$

For $n=k+1$ ,

\begin{aligned} f^{(k+1)} (x) &= \frac{d}{dx} f^{(k)} (x) \\ &= \frac{d}{dx} \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} e^{e^{e^x}} (e^{e^x})^i (e^x)^j \\ &= \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} \Big[ \frac{d}{dx} (e^{e^{e^x}}) \Big] (e^{e^x})^i (e^x)^j + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} e^{e^{e^x}} \Big[ \frac{d}{dx} ( (e^{e^x})^i) \Big] (e^x)^j + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} e^{e^{e^x}} (e^{e^x})^i \Big[ \frac{d}{dx} ( (e^x)^j ) \Big] \\ &= \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} \Big[ e^{e^{e^x}} e^{e^x} e^x \Big] (e^{e^x})^i (e^x)^j + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} e^{e^{e^x}} \Big[ i (e^{e^x})^i e^x\Big] (e^x)^j + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} e^{e^{e^x}} (e^{e^x})^i \Big[ j (e^x)^j \Big] \\ &= \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} c_{kij} e^{e^{e^x}} (e^{e^x})^{i+1} (e^x)^{j+1} + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} i c_{kij} e^{e^{e^x}} (e^{e^x})^i (e^x)^{j+1} + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} j c_{kij} e^{e^{e^x}} (e^{e^x})^i (e^x)^j \\ &= \displaystyle \sum_{\substack{2 \leq i \leq k+1 \\ 2 \leq j \leq k+1 }} c_{k,i-1,j-1} e^{e^{e^x}} (e^{e^x})^i (e^x)^j + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 2 \leq j \leq k+1 }} i c_{ki,j-1} e^{e^{e^x}} (e^{e^x})^i (e^x)^j + \displaystyle \sum_{\substack{1 \leq i \leq k \\ 1 \leq j \leq k }} j c_{kij} e^{e^{e^x}} (e^{e^x})^i (e^x)^j \\ &= \displaystyle \sum_{\substack{1 \leq i \leq k+1 \\ 1 \leq j \leq k+1 }} c_{k+1,ij} e^{e^{e^x}} (e^{e^x})^i (e^x)^j \\ \end{aligned}

The coefficients $c_{k+1,ij}$ are

$\begin{array}{lclcl} c_{k+1,ij} = 0 & \text{for} & i = k+1 & \text{and} & j = 1 \\ c_{k+1,ij} = i c_{ki,j-1} & \text{for} & i = 1 & \text{and} & j = k+1 \\ c_{k+1,ij} = j c_{kij} & \text{for} &1 \leq i \leq k & \text{and} & j = 1 \\ c_{k+1,ij} = c_{k,i-1,j-1}& \text{for} & i = k+1 & \text{and} & 2 \leq j \leq k+1 \\ c_{k+1,ij} = i c_{ki,j-1} + j c_{kij} & \text{for} & i = 1 & \text{and} & 2 \leq j \leq k \\ c_{k+1,ij} = c_{k,i-1,j-1} + i c_{ki,j-1} & \text{for} & 2 \leq i \leq k & \text{and} & j = k+1 \\ c_{k+1,ij} = c_{k,i-1,j-1} + i c_{ki,j-1} + j c_{kij} & \text{for} & 2 \leq i \leq k & \text{and} & 2 \leq j \leq k \\ \end{array}$

Step 3:

Let $y = (e^{e^{e^x}})^{-1} (e^{e^x})^{-1} (e^x)^{-1}$

Prove that $\displaystyle \lim_{x \rightarrow \infty} f^{(n)} (x) \cdot y^n =\left\{ \begin{array}{rl} 1 & \text{if }n=1 \\ 0 & \text{if }n \geq 2 \end{array} \right.$

For $n = 1$ ,

$\begin{aligned} \displaystyle & \lim_{x \rightarrow \infty} f^{(1)} (x) \cdot y^1 \\ &= \displaystyle \lim_{x \rightarrow \infty} e^{e^{e^x}} e^{e^x} e^x \cdot (e^{e^{e^x}})^{-1} (e^{e^x})^{-1} (e^x)^{-1} \\ &= \displaystyle \lim_{x \rightarrow \infty} 1 \\ &= 1 \end{aligned}$

For $n \geq 2$ ,

\begin{aligned} \displaystyle & \lim_{x \rightarrow \infty} f^{(n)} (x) \cdot y^n \\ &= \displaystyle \lim_{x \rightarrow \infty} \sum_{\substack{1 \leq i \leq n \\ 1 \leq j \leq n }} c_{nij} e^{e^{e^x}} (e^{e^x})^i (e^x)^j \cdot (e^{e^{e^x}})^{-n} (e^{e^x})^{-n} (e^x)^{-n} \\ &= \displaystyle \lim_{x \rightarrow \infty} \sum_{\substack{1 \leq i \leq n \\ 1 \leq j \leq n }} c_{nij} (e^{e^{e^x}})^{-(n-1)} (e^{e^x})^{-(n-i)} (e^x)^{-(n-j)} \\ &= 0 \end{aligned}

Step 4:

The value of the limit is

$\begin{aligned} \displaystyle & \lim_{x \rightarrow \infty} \Big[ f \Big(x + e^{-(a + x + e^x + e^{e^x} )} \Big) - f(x) \Big] \\ &= \displaystyle \lim_{x \rightarrow \infty} \Big[ f \Big( x + e^{-a} e^{-x} e^{-e^x} e^{-e^{e^x}} \Big) - f(x) \Big] \\ &= \displaystyle \lim_{x \rightarrow \infty} \Big[ f \Big( x + e^{-a} (e^x)^{-1} (e^{e^x})^{-1} (e^{e^{e^x}})^{-1} \Big) - f(x) \Big] \\ &= \displaystyle \lim_{x \rightarrow \infty} \Big[ f ( x + e^{-a} y ) - f(x) \Big] \\ &= \displaystyle \lim_{x \rightarrow \infty} \Big[ f(x) + f^{\prime}(x) (e^{-a} y) + \sum_{n=2}^{\infty}\Big[ \frac{f^{(n)}(x)}{n!} (e^{-a} y)^n \Big] - f(x) \Big] \\ &= \displaystyle \lim_{x \rightarrow \infty} \Big[ f^{\prime}(x) (e^{-a} y) \Big] + \displaystyle \lim_{x \rightarrow \infty} \sum_{n=2}^{\infty}\Big[ \frac{f^{(n)}(x)}{n!} (e^{-a} y)^n \Big] \\ &= e^{-a} \displaystyle \lim_{x \rightarrow \infty} \Big[ f^{\prime}(x) \cdot y \Big] + \displaystyle \lim_{x \rightarrow \infty} \sum_{n=2}^{\infty}\Big[ \frac{e^{-na}}{n!} f^{(n)}(x) \cdot y^n \Big] \\ &= e^{-a} + \displaystyle \sum_{n=2}^{\infty}\Big[ \frac{e^{-na}}{n!} \lim_{x \rightarrow \infty} \big( f^{(n)}(x) \cdot y^n \big) \Big] \\ &= e^{-a} + \displaystyle \sum_{n=2}^{\infty}\Big[ \frac{e^{-na}}{n!} \cdot 0 \Big] \\ &= e^{-a} + \displaystyle \sum_{n=2}^{\infty}0 \\ &= e^{-a} + 0 \\ &= e^{-a} \end{aligned}$