We give two solutions. The first is a formal calculation. The second is informal, but more informative.
Formal verification: Let $X$ have exponential distribution with parameter $\lambda$. Let $k$ be a positive constant, and let $Y=kX$. The cumulative distribution function $F_Y(y)$ of $Y$, for $y \ge 0$, is given by
$$F_Y(y)=P(Y \le y)=P(X \le y/k)=F_X(y/k),$$
where $F_X$ is the cumulative distribution function of $X$.
Now there are several ways to find the density function $f_Y(y)$ of $Y$: (i) We could find $F_X(x)$ by integrating, substitute $y/k$ for $x$, and differentiate with respect to $y$; (ii) We could look up $F_X(x)$ and then proceed as in (i); (iii) Or else we can use the Fundamental Theorem of Calculus.
Let $x=y/k$. By the Chain Rule,
$$\frac{dF_X(y/k)}{dy}=\frac{dx}{dy}\frac{dF_X(x)}{dx}=\frac{1}{k}f_X(x)=\frac{1}{k}\lambda e^{-\lambda x}=\frac{\lambda}{k}e^{-\lambda y/k}=\lambda'e^{-\lambda' y},$$
where $\lambda'=\lambda/k$. That is what we wanted to show.
Note: The approach (iii) is in general the easiest, so that's what we used. It may look more difficult than approach (i), but that's because $\lambda e^{-\lambda x}$ is exceptionally easy to integrate. If we are working with the normal instead of the exponential, we can't even start on (i), while (iii) goes through just as easily as for the exponential.
Informal justification: The random variable $X$ might describe the lifetime, in days, of an atom of a certain radioactive substance $S$. Recall that the mean lifetime is $1/\lambda$.
On Planet $K$ far far away, the length of the day is one-third the length of an Earth day. So, on Planet $K$, and measuring time in $K$-days, substance $S$ will have mean lifetime $3/\lambda$ $K$-days. It follows that the lifetime $Y$ of an atom of substance $S$, as measured in $K$-days, has exponential distribution with parameter $\lambda/3$.
Similarly, if the length of the day on Planet $K$ is $1/k$ times an Earth day,
then on Planet $K$, measuring time in $K$-days, substance $S$ has mean lifetime $k/\lambda$. So the lifetime $Y$ of an atom of $S$, measured in $K$-days, has exponential distribution with parameter $\lambda/k$.
Everything is correct except for your last line. Instead:
$$\sum_{n=1}^{\infty}\prod_{k=1}^{n}E\left[X_1\right]=\sum_{n=1}^{\infty}\prod_{k=1}^{n}1/\lambda \color{red}=\sum_{n=1}^{\infty} \frac{1}{{\lambda^n}}= \frac{1}{\lambda - 1}$$
where in the last step we used the formula for a geometric series.
Best Answer
A few notes that might be helpful.
Suppose $0 < a_0 < a_1 < \cdots$ and that $X$ has density \begin{equation*} f(x) = \sum_{m=0}^{M-1} C_m e^{-a_m x} \end{equation*} on $x \geq 0$, where the $C_m$ are not necessarily nonnegative, and that $Y$ is independent and distributed as $a_M e^{-a_M y}$, $y \geq 0$. Then $Z = X + Y$ has density \begin{align*} f_Z(z) &= \int_0^z f(x) a_M e^{-a_M(z - x)}\, dx \\ &= \int_0^z\, \sum_{m=0}^{M-1}a_M C_m e^{-a_mx}e^{-a_Mz + a_Mx}\, dx \\ &= \sum_{m=0}^{M-1}a_M C_m e^{-a_Mz}\int_0^z\, e^{-a_mx}e^{a_Mx}\, dx \\ &= \sum_{m=0}^{M-1}a_M C_m e^{-a_Mz}\int_0^z\, e^{(a_M - a_m)x}\, dx \\ &= \sum_{m=0}^{M-1}\frac{a_M C_m}{a_M - a_m} e^{-a_Mz} (e^{(a_M - a_m)z} - 1) \\ &= \sum_{m=0}^{M-1}\frac{a_M C_m}{a_M - a_m} (e^{-a_mz} - e^{-a_Mz}) \\ &= \sum_{m=0}^{M-1}\frac{a_M C_m}{a_M - a_m} e^{-a_mz} - \bigl( \sum_{m=0}^{M-1}\frac{a_M C_m}{a_M - a_m}\bigr) e^{-a_Mz} \end{align*} which shows $f_Z$ has a form similar to the above: \begin{equation*} f_Z(z) = \sum_{m=0}^M C'_m e^{-a_m z} \end{equation*} with \begin{equation*} C'_m = \frac{a_M C_m}{a_M - a_m} \end{equation*} for $m = 0, \dots, M-1$ and \begin{equation*} C'_M = -\sum_{m=0}^{M-1} \frac{a_M C_m}{a_M - a_m}. \end{equation*} This lends itself to an iteration constructing the limiting density as $M \rightarrow \infty$, at least when it exists.
The original question involves $a_m = 2^m$ so we expect a density of the form \begin{equation*} f_\infty(z) = D_0e^{-z} + D_1e^{-2z} + D_2e^{-4z} + D_3e^{-8z} + \dots \end{equation*} The recursion for the leading coefficient yields $D_0 = 1/ \prod_{k=1}^\infty (1 - 2^{-k}) \approx 3.462746619$.
ADDED
The first few coefficients, computed using the recurrence above, are
\begin{align*} m &\qquad\qquad D_m \\ 0 &\qquad +3.46274662e+00 \\ 1 &\qquad -6.92549324e+00 \\ 2 &\qquad +4.61699549e+00 \\ 3 &\qquad -1.31914157e+00 \\ 4 &\qquad +1.75885543e-01 \\ 5 &\qquad -1.13474544e-02 \\ 6 &\qquad +3.60236646e-04 \\ 7 &\qquad -5.67301805e-06 \\ 8 &\qquad +4.44942599e-08 \\ 9 &\qquad -1.74147636e-10 \end{align*}